Variable Acceleration Motion

  • 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 Examples

Solved Example:

28-4-01

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

Solution:
\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)

Solution:
\[\dfrac{dv}{dt} = - 2.5 \sqrt{v}\] \[\dfrac{dv}{\sqrt{v}} = - 2.5 dt\] 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.

28-4-03



Solution:
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:

Solution:
\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