Learning Objectives:

• To define and calculate the instantaneous velocity and speed.

• To differentiate between average and instantaneous velocity.

• To calculate the average acceleration and determine its direction.

• To calculate the instantaneous acceleration from position function or velocity function and determine its direction.

The instantaneous velocity is obtained as follows: \[\vec{v} = \lim_{\Delta t \to 0}\frac{\Delta s}{\Delta t}\] Or, if s = f(t) then v = \(\dfrac{d}{dt}f(t)\) = f’(t)
The instantaneous acceleration is obtained as follows: \[\vec{a} = \lim_{\Delta t \to 0}\frac{\Delta v}{\Delta t}\] Or, if v = g(t) then a = \(\dfrac{d}{dt}g(t)\) = \(\dfrac{d^2}{dt^2}f(t)\)

Solved Example: 28-1-01

A particle moves along a straight line with the equation x = 16t + 4$t^2$ - 3$t^3$ where x is the distance in m and t is the time in second. Compute the acceleration of the particle after 2 seconds.

A. -28 $m/s^2$

B. -30 $m/s^2$

C. -17 $m/s^2$

D. -24 $m/s^2$

\begin{align*} x &= 16t + 4t^2 - 3t^3\\ \dot{x} &= 16 + 4(2t) - 3(3t^2)\\ \ddot{x} &= 4(2) - 3(3)(2t) = 8 - 18t \end{align*} At t= 2 sec, $\ddot{x} = 8 - 18(2) = -28\ m/s^2$

Correct Answer: A

Solved Example: 28-1-02

A particle moves along a straight line such that distance (x) traversed in 't' seconds is given by $x= t^2*(t-4)$, the acceleration of the particle will be given by the equation:

A. $\ddot{x}= t^2*(t-4)$

B. $\ddot{x}= 2t$

C. $\ddot{x}= 6t-8$

D. $\ddot{x}= 3t^2- 8t$

The acceleration can be found out by differentiating the displacement function twice with respect to time. \[x= t^2*(t-4) = t^3 - 4t^2\] Differentiating, $v = 3t^2 - 8t$
Differentiating one more time, $a = 6t - 8$

Correct Answer: C

Solved Example: 28-1-03

The motion of a particle is defined by the relation x = (1/3)$t^3$ - 3$t^2$ + 8t + 2 where x is the distance in meters and is the time in seconds. What is the time when the velocity is zero?

A. 2 seconds

B. 3 seconds

C. 5 seconds

D. 7 seconds

\begin{align*} x &= \dfrac{1}{3}t^3 - 3t^2 + 8t + 2\\ \dot{x} &= \dfrac{1}{3}\times 3 t^2 - 3\times 2t + 8 = t^2 - 6t + 8 = (t-4)(t-2) \end{align*} Hence, velocity is zero when t = 2 or t = 4

Correct Answer: A

Solved Example: 28-1-04

An equation x = $t^3$ - $2t$ denoted the relationship between displacement and time. At t = 4 sec. acceleration is given by:

A. 24 units

B. 26 units

C. 12 units

D. 22 units

Velocity v= $\dfrac{dx}{dt}$ = $3t^2 -2$, acceleration a= $\dfrac{dv}{dt} = 6t$
At t = 4 sec, a = 24 units

Correct Answer: A

Solved Example: 28-1-05

The rate of change of displacement is called ________. (RRC Group D Sep 2018 Shift I)

A. Distance

B. Velocity

C. Acceleration

D. Force

Correct Answer: B

Solved Example: 28-1-06

During uniform motion of an object along a straight line, its _______ remains constant with time. (SSC CHSL March 2018 Shift III)

A. Acceleration

B. Deceleration

C. Force

D. Velocity

Since the motion is along a straight line, the direction of motion is NOT changing.
Also, since the motion is uniform, the distance convered in equal intervals is same too.
That means, the magnitude as well as the direction of the displacement is NOT changing, hence the velocity is constant.

Correct Answer: D

Solved Example: 28-1-07

If a body after travelling some distance comes back to its starting point then: (UP TGT Science 2011)

A. Average velocity is zero.

B. Average speed is zero.

C. Distance traveled is zero.

D. All the above.

Average velocity is given by: \[\bar{v} = \dfrac{\Delta x}{\Delta t}\] Since $\Delta x = 0$ the average velocity is zero.

In scalar terms, Distance is related to speed.
In vector terms, displacement is related to velocity (and eventually to accelration)

Displacement considers only initial and final positions. Distance considers the path too.
Hence in this case, the displacement is zero, however distance traveled is NOT zero. So C is INCORRECT.

Since distance travelled in not zero, hence average speed is also NOT zero, hence option B is also INCORRECT.

Correct Answer: A

Learning Objectives:

• Calculate required parameters - such as curvature and radius of curvature - for motion along plane curved path.

• Find tangential and normal components of acceleration

The tangential component of acceleration is: \(a_{t} = \dot{v}\)
and the normal component is given by: \(a_{n} = \dfrac{v^{2}}{\rho }\)
The magnitude of the acceleration is: \[a = \sqrt{{a_t}^2 + {a_n}^2}\] In vector notation,

\[\vec{v} = v(t) \hat{e_t}\]

\[\vec{a} = a(t) \hat{e_t} + \dfrac{v^{2}}{\rho} \hat{e_n}\]

\[v = v(t) \mathbf{e_{t}}\] \[\vec{a} = a(t) \mathbf{e_{t}} + \left ( \dfrac{v_{t}^{2}}{\rho } \right )\mathbf{e_{n}}\] where \(\displaystyle \large \rho\) = instantaneous radius of curvature.

Solved Example: 28-2-01

A body is moving along a circular path with variable speed. It has:

A. A radial acceleration

B. A tangential acceleration

C. Zero acceleration

D. Both tangential and radial accelerations

Since the body has variable acceleration, it has tangential acceleration. Also, since the body is moving on a circular path, that means it is continuously changing its direction, so it has centripetal acceleration acting towards the center of curvature. That means it has both tangential and radial accelerations.

Correct Answer: D

Solved Example: 28-2-02

A particle is under uniform circular motion. Which is the wrong statement regarding its motion?

A. The velocity vector is tangential to the circle

B. The acceleration vector is tangential to the circle

C. The acceleration vector is directed towards the centre of the circle

D. The velocity and acceleration vectors are perpendicular to each other

Since the particle is under UNIFORM circular motion, it will have no tangential acceleration, only radial (normal) acceleration will be present. Its velocity will be towards the tangent. Hence velocity and acceleration will be perpendicular to each other. (valid only for UNIFORM circular motion.) Let us see each statement individually.

• The velocity vector is tangential to the circle: CORRECT

• The acceleration vector is tangential to the circle. WRONG since the motion is uniform, there is NO tangential acceleration, but it will have a centripetal acceleration which is towards the center of the curvature. So the acceleration vector is NORMAL or RADIAL.

• The acceleration vector is directed towards the centre of the circle. CORRECT

• The velocity and acceleration vectors are perpendicular to each other. CORRECT

Correct Answer: B

Solved Example: 28-2-03

A particle revolves round a circular path with uniform angular velocity. The acceleration of the particle is:

A. Along the circumference of the circle

B. Along the tangent

C. Along the radius

D. Zero

The key word is UNIFORM, hence tangential acceleration is absent, only radial acceleration will be there.

Correct Answer: C

Solved Example: 28-2-04

A car is moving with speed 30 m/s on a circular path of radius 500 m. Its speed is increasing at the rate of 2 m/s$^2$. The acceleration of the car is:

A. 2 $m/s^2$

B. 1.8 $m/s^2$

C. 9.8 $m/s^2$

D. 2.7 $m/s^2$

\[a_{t}=2m/s^{2}\] \[a_{n}=\dfrac {v^{2}}{r}=\dfrac {30^{2}}{500}=1.8m/s^{2}\] \[a=\sqrt {a^{2}_{n}+a^{2}_{t}}=\sqrt {2^{2}+1.8^{2}}=2.69\ m/s^{2}\]

Correct Answer: D

Solved Example: 28-2-05

If cycle wheel of radius 0.4m completes 1 rev in 1 sec, then acceleration of the cycle is:

A. 1.6 $\pi^2 m/s^2$

B. 0.4 $\pi m/s^2$

C. 0.8 $\pi m/s^2$

D. 0.4 $\pi^2 m/s^2$

\[a=\dfrac {v^{2}}{r}=r\omega ^{2}\] \[\omega =\dfrac {1\ rev}{1\ sec}=2\pi\ rad/s\] \[a=\left( 0.4\right) \left( 2\pi \right) ^{2}=1.6\pi ^{2}m/s^{2}\]

Correct Answer: A

Solved Example: 28-2-06

A particle originally at rest at the highest point on a smooth vertical circle is slightly displaced. It will leave the circle at a vertical distance h below the highest point such that:

A. h= R

B. h= $\dfrac{R}{3}$

C. h = $\dfrac{R}{2}$

D. h = $\dfrac{2R}{3}$

Correct Answer: B

velocity at any height h from the top is given by: $v = \sqrt{2gh}$
\[\mathrm{centrifugal\ force} = \dfrac{mv^2}{R} = \dfrac{2mgh}{R}\] Refer to the free body diagram as shown, \[\dfrac{mv^2}{R} + N = mg \sin \theta\] The particle will leave the circle when normal reaction N = 0
\begin{align*} \dfrac{2gh}{R} &= g \sin \theta\\ \dfrac{2gh}{R} &= g \left(\dfrac{R-h}{R}\right)\\ 2h &= (R-h)\\ 3h &= R\ \ Or, \ h = \dfrac{R}{3} \end{align*}

Learning Objectives:

• Draw, analyze and interpret position-time, displacement-time, velocity-time and acceleration-time graphs.

• Solve simple problems involving uniform motion and uniformly accelerated motion by using appropriate equations.

\[\begin{aligned} v &= u + at\\ s &= ut + \dfrac{1}{2} at^2\\ v^2 &= u^2 + 2as\end{aligned}\] where, u= Initial velocity, v= Final velocity
t= time
s= displacement
a= acceleration
In case of motion under gravity, which falls under constant acceleration motion, a = -9.81 \(m/s^2\)

Solved Example: 28-3-01

Water drops from a faucet at the rate of 4 drops per second. What is the distance between two successive drops 1 second after the first drop has fallen.

A. 1.62 m

B. 2.51 m

C. 2.15 m

D. 2.87 m

Time interval between drops = 0.25 s
Drop A travels for 1 seconds.
Drop B starts journey after 0.25 sec after A has begun falling. \[\begin{aligned} u &=0\ m/s,\\ t &= 1\ s,\\ a &= 9.81\ m/s^2 \end{aligned}\] Distance traveled by drop A \[s = ut + \dfrac{1}{2}at^2 =\dfrac{1}{2}*9.81*1*1 =4.905\ m\] Now, drop B travels for 1 - 0.25 = 0.75 sec \[s = \dfrac{1}{2}*9.81*0.75*0.75 =2.76\ m\] Distance between drop A and drop B = 4.905-2.76= 2.15 m

Correct Answer: C

Solved Example: 28-3-02

A motorist is traveling at 90 kmph. When he observes a traffic signal 250m ahead of him turns red. The traffic signal time to stay red is 12 sec. If the motorist wishes to pass the signal without stopping as it turns green, determine the required uniform deceleration of the motor.

A. 0.3154 $m/s^2$

B. 0.9130 $m/s^2$

C. 0.3711 $m/s^2$

D. 0.6944 $m/s^2$

$u = 90\ \mathrm{kmph} = 90 \times \dfrac{5}{18}$ = 25 m/s
s = 250 m,
t =12 sec
Using, \begin{align*} s &= ut + \dfrac{1}{2} at^2\\ 250 &= 25 \times 12 + \dfrac{1}{2} a \times 12^2\\ a &= -0.6944\ m/s^2 \end{align*}

Correct Answer: D

Solved Example: 28-3-03

Two cars A and B traveling in the same direction and stopped at a highway traffic sign. As the signal turns green car A accelerates at constant rate of 1 $m/s^2$. Two seconds later the second car B accelerates at constant rate of 1.3 $m/s^2$. When will the second car B overtakes the first car A?

A. 16.27 s

B. 30.45 s

C. 20.32 s

D. 10.45 s

Distance travelled by A in 't' sec = Distance travelled by B in (t-2) sec. \[\dfrac {1}{2}a_{1}t^{2}=\dfrac {1}{2}a_{2}\left( t-2\right) ^{2}\] \[\left( 1\right) \left( t\right) ^{2}=\left( 1.3\right) \left( t-2\right) ^{2}\] \[t^{2}=1.3\left( t^{2}-4t+4\right)\] \[0.3t^{2}-5.2t+5.2=0\] \[t=1.07 \ s \ or\ t=16.27 \ s\] But t cannot be less than 2 seconds. Hence, t= 16.27 sec

Correct Answer: A

Solved Example: 28-3-04

A car starts from rest and has a constant acceleration of 3 $m/s^2$. Find the average velocity during the first 10 seconds of motion.

A. 13 m/s

B. 15 m/s

C. 14 m/s

D. 20 m/s

\[v=u+at =\left( 3\right) \left( 10\right) =30\ m/s\] Average velocity is given by- \[=\dfrac {0+30}{2}=15\ m/s\]

Correct Answer: B

Solved Example: 28-3-05

A ball of mass 0.1 kg, initially at rest, is dropped from height of 1 m. Ball hits ground and bounces off ground. Upon impact with the ground, the velocity reduces by 20%. The height (in m) to which the ball will rise is: (GATE ME 2015)

A. 0.46

B. 0.55

C. 0.59

D. 0.64

Velocity just before impact with the ground,
\begin{align*} v^2 &= u^2 + 2gs\\ v^2 &= 0 + 2\times 9.81 \times 1\\ v &= 4.43\ m/s \end{align*} It is given that the velocity after impact is 80%
\[v_{\mathrm{after\ impact}} = 0.8 \times 4.43 = 3.54\ m/s\] This is now taken as u for return (upward) motion. The final velocity, at the topmost point, will be zero.
\begin{align*} v^2 &= u^2 + 2gs\\ 0 &= 3.54^2 + 2 (-9.81) s\\ s &=0.64\ m \end{align*}

Correct Answer: D

Solved Example: 28-3-06

A ball is thrown vertically into the air at 120 m/s. After 3 seconds, another ball is thrown vertically. What is the velocity the second ball must have, in m/s, so that it collides the first ball at 100 m from the ground?

A. 105.89

B. 107.72

C. 108.12

D. 110.72

First ball travels 100m distance when thrown with an initial velocity of 120 m/s. Let's find the time of travel. \begin{align*} s&=ut+\dfrac{1}{2}at^{2}\\ 100&=120t+\dfrac{1}{2}\left(-9.81\right) t^{2}\\ t&=23.626\ \mathrm{sec} \end{align*} which means the other ball travels 23.626 - 3 =20.626 seconds. We will find its initial velocity such that it crosses the first ball 100 m above the ground. Let's assume the velocity of initial is u$_2$ . \begin{align*} s&=u_{2}t +\dfrac{1}{2}at^{2}\\ 100&=u_{2}\left(20.626\right) +\dfrac{1}{2}\left(-9.81\right) \left( 20.626\right)^{2}\\ u_{2}&=105.9\ \mathrm{m/s} \end{align*}

Correct Answer: A

Solved Example: 28-3-07

A ball is dropped from a height of 60 meters above ground. How long does it take to hit the ground?

A. 2.1 s

B. 3.5 s

C. 5.5 s

D. 1.3 s

Here, s = -60 m, a = -9.81 m/s$^2$, u = 0 m/s \begin{align*} s &= ut + \dfrac{1}{2}at^2 \\ -60 &= 0 + \dfrac{1}{2}(-9.81)t^2 \\ t &= 3.5\ \mathrm{sec} \end{align*}

Correct Answer: B

Solved Example: 28-3-08

A ball is thrown vertically upward from the ground and a student gazing out of the window sees it moving upward pass him at 5 m/s. The window is 10 m above the ground. How high does the ball go above the ground?

A. 15.25 m

B. 14.87 m

C. 9.97 m

D. 11.28 m

For motion from ground to window: u = ?, v = 5 m/s, s = 10 m
\begin{align*} v^{2} &=u^{2}+2gs\\ 25 &= u^{2}+2\left(-9.81\right) \left(10\right)\\ u &= 14.87\ m/s \end{align*} For motion from ground to the topmost point in air: u = 14.87 m/s, v = 0, s = ? \begin{align*} v^{2} &=u^{2}+2gs\\ 0 &=\left(14.87\right) ^{2}+2\left(-9.81\right) s\\ s &=11.27\ m \end{align*}

Correct Answer: D

Solved Example: 28-3-09

A ball thrown vertically upward with an initial velocity of 3 m/s from the window of a tall building. The ball strikes the sidewalk at the ground level 4 seconds later. Determine the velocity with which the ball strikes the ground.

A. 39.25 m/s

B. 38.50 m/s

C. 37.75 m/s

D. 36.24 m/s

We will consider the motion in two parts: One, above the building and second, below the building.

Part I: (Motion above the building, back to original throwing point)
s = 0
u = 3 m/s
t = ?
a = -9.81 m/s$^2$

Using, \[s = ut + \dfrac{1}{2} at^2\] t = 0.6116 sec

Part II: Motion below the building.
That leaves 4 sec - 0.6116 = 3.3884 seconds for this motion

using, \[v = u + ut\] \[v = 3 + 9.81 (3.3884) = 36.24 m/s\]

Correct Answer: D

Solved Example: 28-3-10

A car accelerates from rest at a constant rate '$\alpha$' for some time, after which it decelerates at a constant rate '$\beta$' and comes to rest. If the total time elapsed is t, then the maximum velocity acquired by the car is: (NEET 1994, JEE(Mains) 2021-Physics-Section A- Q12)

A. $\left(\dfrac{\alpha^2 + \beta^2}{\alpha \beta}\right)t$

B. $\left(\dfrac{\alpha^2 - \beta^2}{\alpha \beta}\right)t$

C. $\dfrac{(\alpha + \beta)t}{\alpha \beta}$

D. $\dfrac{\alpha \beta t}{(\alpha + \beta)}$

Let the car accelerate at rate $\alpha$ for time t$_1$ then maximum velocity attained, \[v_{max}=0 + \alpha t_1 = \alpha t_1\] Now, the car decelerates at a rate $\beta$ for time (t-t$_1$) and finally comes to rest. Then, \begin{align*} \dfrac{v_{max}- 0 }{t - t_1} &= \beta\\ \dfrac{v_{max}}{ t- \left(\dfrac{v_{max}}{\alpha}\right)} &= \beta \\ \dfrac{\alpha v_{max}}{\alpha t - v_{max}} &= \beta \\ \alpha v_{max} &= \alpha \beta t - \beta v_{max}\\ v_{max} (\alpha + \beta ) &= \alpha \beta t\\ v_{max} &= \left( \dfrac{\alpha \beta }{\alpha + \beta}\right)t \end{align*}

Correct Answer: D

Learning Objectives:

• Integrate acceleration function to find velocity and position vectors or displacement.

Case I:
If a = f(t) then integrate twice to get displacement function. Do not forget constant of integration which can be evaluated by applying boundary conditions.

\[v(t) = \int_{t_0}^{t_1} a(t)dt + v_0\]

\[s(t) = \int_{t_0}^{t_1} v(t)dt + s_0\]

Case II:
If a = f(v) use a = v\(\dfrac{dv}{ds}\) bring all similar terms on one side and integrate to get velocity function.
Case III:
If a = f(s) again use a = v\(\dfrac{dv}{ds}\) bring all similar terms on one side and integrate to get velocity function.

Solved Example: 28-4-01

The velocity time relationship is described by the equation $v = P + Qt^2$. The body is travelling with:

A. Uniform retardation

B. Zero acceleration

C. Uniform acceleration

D. Non-uniform acceleration

\begin{align*} v &= P + Qt^2\\ a &= \dfrac{dv}{dt} = 2Qt \end{align*} Hence acceleration depends upon time and hence it is not constant.(non-uniform).

Correct Answer: D

Solved Example: 28-4-02

An object, moving with a speed of 6.25 $ms^{-1}$, is decelerated at a rate given by: \[ \dfrac{dv}{dt} = -2.5 \sqrt{v}\] where v is the instantaneous speed. The time taken by the object, to come to rest, would be: (JEE Main Physics 2011)

A. 2 s

B. 4 s

C. 8 s

D. 1 s

\[\dfrac{dv}{dt} = - 2.5 \sqrt{v}\] \[\dfrac{dv}{\sqrt{v}} = - 2.5 dt\] Integrating both sides, \[2 \sqrt{v} = -2.5 t + C \] Applying boundary condition,
At t = 0, v = 6.25 \[C = 5\] Substituting, \[2 \sqrt{v} = -2.5 t + 5\] when v = 0 , t= ? \[0 = -2.5 t + 5\] \[t = 2\ sec\]

Correct Answer: A

Solved Example: 28-4-03

A particle of mass 1 kg is acted upon by a force which varies as shown in the figure. If initial velocity of the particle is 10 m/s, determine what is the maximum velocity attained by the particle.

A. 110 m/s

B. 100 m/s

C. 90m/s

D. 80 m/s

The first part of motion is under variable acceleration as the force itself is variable. \[v_A = v_0 + \mathrm{Area\ of\ triangle} = 10 + \dfrac{1}{2} \times 10 \times 20 = 110\ m/s\] The next part of the motion is under negative force.
Hence, the velocity will go on decreasing.
Hence, maximum velocity will NOT be affected.

Correct Answer: A

Solved Example: 28-4-04

The acceleration of a particle is increasing linearly with time t as bt. The particle starts from the origin with an initial velocity v$_0$ The distance travelled by the particle in time t will be:

A. v$_0$t + $\dfrac{1}{3}$bt$^2$

B. v$_0$t + $\dfrac{1}{3}$bt$^3$

C. v$_0$t + $\dfrac{1}{6}$bt$^3$

D. v$_0$t + $\dfrac{1}{2}$bt$^2$

\begin{align*} a&=bt\\ \dfrac {dv}{dt}&=bt\\ dv&=bt\ dt\\ v&=\dfrac {bt^{2}}{2}+c_{1}\\ At\ t=0,v=v_{0}\\ v_{0}&=0+c_{1}\\ c_{1}&=v_{0} \end{align*} \begin{align*} v&=v_{0}+\dfrac {bt^{2}}{2}\\ \dfrac {ds}{dt}&=v_{0}+\dfrac {bt^{2}}{2}\\ ds&=\left[ v_{0}+\dfrac {bt^{2}}{2}\right] dt\\ \int ds&=\int \left[ v_{0}+\dfrac {bt^{2}}{2}\right] dt\\ s&=v_{0}t+\dfrac {bt^{3}}{6}+c_{2}\\ s|_{t_{1}\rightarrow t_{2}}&=v_{0}t+\dfrac {bt^{3}}{6} \end{align*}

Correct Answer: C

Learning Objectives:

• Define relative velocity and explain examples where the concept of relative velocity is useful.

• Calculate relative velocity graphically by applying the concepts of vectors.

When velocities are along a straight line in the same or opposite direction and all have the same reference, we can find relative velocities by vector subtraction. \[\begin{aligned} v_{A/B} &= v_A - v_B\\ v_{A/B} &= - v_{B/A}\\ v_{A/C} &= v_{A/B} + v_{B/C}\end{aligned}\] A person walks on a treadmill moving at certain velocity and remains at the same place in the gym. The person’s velocity relative to the gym floor is zero.

Solved Example: 28-5-01

Rain is falling vertically with the speed of 30 m/s. A man rides a bicycle with the speed of 10 m/s in east to west direction. What is the direction in which he should hold the umbrella?

A. $\cos^{-1}\left(\dfrac{1}{3}\right)$

B. $\sin^{-1}\left(\dfrac{1}{3}\right)$

C. $\tan^{-1}\left(\dfrac{2}{3}\right)$

D. $\tan^{-1}\left(\dfrac{1}{3}\right)$

Correct Answer: D

Solved Example: 28-5-02

A river flows at 2 m/s. The man can swim in still water with velocity of 4 m/s. The angle at which man must swim with the banks so that he crosses the river in the shortest distance is:

A. 30$^\circ$ upstream

B. 30$^\circ$ downstream

C. 60$^\circ$ upstream

D. 60$^\circ$ downstream

Hint: Draw the velocity triangle as a right angled triangle, with hypotenuse as 4m and horizontal side as 2m.

Correct Answer: D

Learning Objectives:

• Learn the relationships between linear and angular distance and velocity for a body in rotation.

Uniform Circular Motion is a movement of an object along the circumference of a circle with uniform angular velocity.
Arc Length \(s = r  \theta\)
Tangential velocity \( v_{t} = r\omega\)
Tangential acceleration \( a_{t} = r\alpha\)
For uniform circular motion, \(\alpha\) = 0, so there is no tangential acceleration.
Normal acceleration \( a_{n} = r \omega ^{2} = \dfrac{v^2}{r}\)

Solved Example: 28-6-01

A particle is rotating about a point with constant angular velocity of 100 rpm. In 2.5 seconds, it will rotate through an angular displacement of:

A. 26.18 rad

B. 34.11 rad

C. 43.71 rad

D. 55.09 rad

\[\omega = \dfrac{2 \pi N}{60}= \dfrac{2 \pi (100)}{60}= 10.47\ rad/s\] \[\theta = \omega.t= 10.47 \times 2.5 = 26.18\ rad\]

Correct Answer: A

Solved Example: 28-6-02

A highway curve has a super elevation of 7$^\circ$. What is the radius of the curve such that there will be no lateral pressure between the tires and the roadway at a speed of 64.36 kmph? (Philippines CE Board Nov 1998)

A. 265.71 m

B. 438.34 m

C. 345.34 m

D. 330.78 m

\[v = 64.36 \times \dfrac{1000}{3600} = 17.87\ \mathrm{m/s}\] \begin{align*} \tan \theta &= \dfrac{v^2}{rg}\\ r &= \dfrac{v^2}{g \tan \theta}\\ r &= \dfrac{(17.87)^2}{9.81 \tan 7^\circ}\\ r &= 265.11\ m \end{align*}

Correct Answer: A

Solved Example: 28-6-03

A girl tied 80 gram toy plane of a string which he rotated to form a vertical circular motion with a diameter a 1000 mm. Compute for the maximum pull exerted on the string by the toy plane if got loose leaving at the bottom of the circle at 25 m/s. (ME Board April 1991)

A. 0.002 kN

B. 0.05 kN

C. 0.2 kN

D. 0.1 kN

\begin{align*} F_C &= \dfrac{mv^2}{r}\\ &= \dfrac{0.08(25)^2}{0.5}\\ &= 100 N\\ &= 0.1 kN \end{align*}

Correct Answer: D

Learning Objectives:

• Calculate the horizontal and vertical components with respect to velocity and position of a projectile at various points along its path.

• Solve problems for projectiles launched horizontally and at various angles to the horizontal to calculate maximum height, range, and overall time of flight of the projectile.

Projectiles are the freely projected particles, which have the combined effect of a vertical as well as a horizontal motion.
Velocity of projection is defined as the velocity with which the particle is projected into space.
Angle of Projection (\(\theta\)) is the angle between the direction of projection and horizontal direction.
Trajectory is the path traced by the projectile.
\[\begin{aligned} v_x &= v_0 \cos \theta \\ v_y &= -gt + v_0 \sin \theta \end{aligned}\]

Time of flight is the time interval during which the projectile is in motion.
Time of flight, \[t = \frac{2v_0 \sin\theta}{g}\]

Maximum height of projectile, \[h_{max} = \dfrac{v_0^2\sin ^2 \theta}{2g}\] Horizontal range is the horizontal distance through which the projectile travels during its flight.
Range = Maximum horizontal distance, \[R= \dfrac{v_0^2}{g} \sin(2 \theta)\]

Solved Example: 28-7-01

The horizontal range and maximum height of a projectile is same. The angle of projection is given by:

A. $\tan \theta$ = 2

B. $\tan \theta$ = 1

C. $\tan \theta$ = 4

D. $\tan \theta$ = 3

$R = \dfrac{v_0^2}{g} \sin(2 \theta)$,

$h_{max} = \dfrac{v_0^2\sin ^2 \theta}{2g}$

But both are given equal, so equating,

Correct Answer: C

Solved Example: 28-7-02

A man aimed his rifle at the bull's eye of a target 50 m away. If the speed of the bullet is 500 m/s, how far below the bull's eye does the bullet strikes the target?

A. 5.0 cm

B. 6.8 cm

C. 5.7 cm

D. 6.0 cm

Horizontal motion: $t=\dfrac {50}{500}=0.1\ s$
Vertical motion: $s =ut +\dfrac{1}{2}at^{2} = \dfrac{1}{2}\left( 9.81\right) \left( 0.1\right)^{2} =0.049\ m\ \simeq 5\ cm$

Correct Answer: A

Solved Example: 28-7-03

A baseball is thrown a horizontal plane following a parabolic path with an initial velocity of 100 m/s at an angle of 30$^\circ$ above the horizontal. Solve the distance from the throwing point that the ball attains its original level.

A. 890 m

B. 883 m

C. 858 m

D. 820 m

Maximum range is given by: \begin{align*} R &= \dfrac{v^2}{g} \sin(2\theta)\\ &= \dfrac{100^2}{9.81} \sin(2 \times 30^\circ)\\ &=883\ m \end{align*}

Correct Answer: B

Solved Example: 28-7-04

An archer must split the apple atop his partner's head from a distance of 30 m. The arrow is horizontal when aimed directly to the apple. At what angle must he aim in order to hit the apple with the arrow traveling at a speed of 35 m/s.

A. 8.35$^\circ$

B. 10.55$^\circ$

C. 3.25$^\circ$

D. 6.95$^\circ$

Maximum range is given by: \begin{align*} R &= \dfrac{v^2}{g} \sin(2\theta)\\ 30 &= \dfrac{35^2}{9.81} \sin(2 \times \theta)\\ \end{align*} Solving, \[\theta = 6.95^\circ\]

Correct Answer: D

Solved Example: 28-7-05

A projectile is fired from a cliff 300 m high with an initial velocity of 400 m/s. If the firing angle is 30$^\circ$ from the horizontal, compute the horizontal range of the projectile.

A. 15.74 km

B. 14.63 km

C. 12.31 km

D. 20.43 km

Vertical motion: $s = ut+ \dfrac{1}{2}at^{2}$
$s = (-300) m,$
$u = 400 \sin 30^\circ\ m/s,$
$a = -9.81\ m/s^2$
Substituting, $4.905t^{2}-200t-300=0, \ \mathrm{Or}\ t=42.22\ sec$
Horizontal motion: \begin{align*} R &=\left( u\cos \theta \right) \cdot t\\ &=\left( 400\cos 30\right) 42\cdot 22\\ &=14626.5\ m\\ &=14.626\ km \end{align*}

Correct Answer: A