Question

In a running competition the runners have to run 100m in a straight line. A runner starts from rest increases his velocity uniformly at the rate of 1m/s up to three quarters of the total displacement and covers the remaining distance with uniform velocity. Find the time taken for 3d quarter and last quarter of the journey.(The instantaneous velocity of the particle is zero. Does it mean that its acceleration is zero?)

Answer 1

In the above question it is given to us that the start from rest increases his velocity uniformly at the rate of 1m/s up to three quarters of the total displacement and covers the remaining distance with uniform velocity. From this information we can say that his acceleration for the first three quarters is $1m{{s}^{-2}}$. The total distance is 100m, therefore we can say that the first three quarters comprise 75m and the last quarter comprises 25m. Using this information we will use these values in Newton’s second kinematic equation to determine the time taken in the respective quarters.

Formula used: $S=Ut+\dfrac{1}{2}a{{t}^{2}}$
${{V}^{2}}-{{U}^{2}}=2aS$

Complete step by step answer:
Let us say a body has an initial velocity ‘U’. If the body accelerates uniformly with acceleration ‘a’, at time t the distance (S)covered by the body is given by,
$S=Ut+\dfrac{1}{2}a{{t}^{2}}$
In the first three quarters i.e. for a distance of 75m, the runner starts from rest and accelerates uniformly with acceleration of $1m{{s}^{-2}}$. Hence using above equation we get the time taken as,
$\begin{aligned} & S=Ut+\dfrac{1}{2}a{{t}^{2}} \\\ & \Rightarrow 75m=0t+\dfrac{1}{2}1m{{s}^{-2}}{{t}^{2}} \\\ & \Rightarrow {{t}^{2}}=150 \\\ & \Rightarrow t=12.2\sec \\\ \end{aligned}$
Let us say at this point of time, the velocity of the runner is V. Using Newton’s third kinematic equation we get,
$\begin{aligned} & {{V}^{2}}-{{U}^{2}}=2aS \\\ & \Rightarrow {{V}^{2}}=2\times 1m{{s}^{-2}}\times 75 \\\ & \Rightarrow V=12.2m{{s}^{-1}} \\\ \end{aligned}$
In the last quarter i.e. for a distance of 25m the runner runs with this speed constantly. Hence using Newton’s second kinematic equation we get time t as,
$\begin{aligned} & S=Ut+\dfrac{1}{2}a{{t}^{2}} \\\ & 25=12.2t+\dfrac{1}{2}(0){{t}^{2}} \\\ & \Rightarrow t=2.04\sec \\\ \end{aligned}$
Therefore the time taken by the runner for the first three quarters by the runner is 12.2sec and that for the last quarter is 2.04sec.

The instantaneous velocity of a particle can be zero but the acceleration need not be zero. We can understand this in the case of the body under the action of gravity as well in the case of action of restoring force.

Note: As the body goes up under the action of gravity, its velocity gradually decreases and reaches zero. But it’s still under the acceleration due to gravity. A similar scenario takes place in case of action of restoring force on a particle. At the extreme position, the particle's velocity is zero but the body remains still under acceleration.