Question

V is the velocity of particle moving on a straight line as a function of time is given as v=5-t then the distance covered in first 10 sec is

Answer 1

The displacement of the particle with respect to time is given by the integral of the velocity.

To find the displacement covered in first 10 seconds, we've to find the integral of velocity between the limits of time 0s and 10s. So,

$\displaystyle s=\int\limits_{0}^{10}v\ dt\\\\\\s=\int\limits_{0}^{10}(5-t)\ dt\\\\\\s=\left[5t-\dfrac{t^2}{2}\right]_{0}^{10}\\\\\\s=\left(5(10)-\dfrac{(10)^2}{2}\right)-\left(5(0)-\dfrac{(0)^2}{2}\right)\\\\\\s=50-\dfrac{100}{2}\\\\\\s=0\ \text{m}$

But the actual displacement is not zero. This is because the area between the graph and time axis is above the time axis between limits 0s and 5s but the same area is below the time axis between limits 5s and 10s, so both are cancelled each other. This is true since the graph is a straight line. We know integration can be used to find the area under the graph.

This is one of the demerit of integration in Physics. If the integral is positive, most of the area under graph is above x - axis (here, time axis). If the integral is zero, an amount of area is above or below the axis for half of the limit chosen, and the same amount is below or above the axis, respectively, between the next half of the limit. If the integral is negative, most of the area under graph is below the axis.

Since the integral of the velocity wrt time between 0s and 10s is zero, what we have to do is to double the integral between the time limit either 0s and 5s, or 5s and 10s. Well, both are same!

So,

$\displaystyle s=2\int\limits_{0}^{5}v\ dt\\\\\\s=2\int\limits_{0}^{5}(5-t)\ dt\\\\\\s=2\left[5t-\dfrac{t^2}{2}\right]_{0}^{5}\\\\\\s=2\left(5(5)-\dfrac{(5)^2}{2}\right)-2\left(5(0)-\dfrac{(0)^2}{2}\right)\\\\\\s=2\left(25-12.5\right)\\\\\\\mathbf{s=25 m}$