Question

A particle starting from rest undergoes acceleration given by $a=|t-2|m/{{s}^{2}}$where t is time in seconds. Velocity of particle after 4 seconds is

& \text{A) 1}m/s \\\ & \text{B) 2}m/s \\\ & \text{C) 8}m/s \\\ & \text{D) }4m/s \\\ \end{aligned}$$

Answer 1

Here we have given a particle which was initially at rest and then undergoes acceleration and acceleration takes place when there is change in velocity. And here we have to calculate the velocity at time 4 seconds. We know that acceleration is the rate of change of velocity with time and the given expression is in terms of time so by integrating it we can find v at particular time t.
Formula used:
$a=\dfrac{dv}{dt}$

Complete answer:
We know that acceleration is given as rate of change of velocity with respect to time which is given as
$a=\dfrac{dv}{dt}$
Where dv represents change in velocity and dt represents change in time.
Now the given value of a is $a=|t-2|$which gives us absolute value that is when $t\le 2$then a will be represented by equation $a=-t+2$and when $t>2$then the equation from a will be $a=t-2$. Hence we can say we have two function of a which are

& a=t-2, t > 2 \\\ & a=-t+2, t\le 2 \\\ \end{aligned}$$ Now by using the formula of acceleration we can write $$\begin{aligned} & |t-2|=\dfrac{dv}{dt} \\\ & dv=|t-2|dt \\\ \end{aligned}$$ Now integrating both sides, the taken limit of t from 0 to 4 initially time will be zero and we have to find velocity at time 4 seconds whereas limit for v will be 0 to v as initially the particle was at rest and assume v is velocity at time 4 seconds. Hence we get $$\int\limits_{0}^{v}{dv}=\int\limits_{0}^{4}{|t-2|dt}$$ But we discuss above that a has two functions therefore for limit 0 to 2 a will be –t+2 and for limit 2 to 4 a will be t-2. $$\begin{aligned} & \int\limits_{0}^{v}{dv}=\int\limits_{0}^{2}{(-t+2)dt}+\int\limits_{2}^{4}{(t-2)dt} \\\ & \left[ v \right]_{0}^{v}=\left[ -\dfrac{{{t}^{2}}}{2}+2t \right]_{0}^{2}+\left[ \dfrac{{{t}^{2}}}{2}-2t \right]_{2}^{4} \\\ & \left( v-0 \right)=\left[ \left( -\dfrac{{{2}^{2}}}{2}+2(2) \right)-0 \right]+\left[ \dfrac{{{4}^{2}}}{2}-2\left( 4 \right)-\dfrac{{{2}^{2}}}{2}+2\left( 2 \right) \right] \\\ & v=\left[ -\dfrac{4}{2}+4 \right]+\left[ \dfrac{16}{2}-8-\dfrac{4}{2}+4 \right] \\\ & v=\left[ -2+4 \right]+\left[ 8-8-2+4 \right] \\\ & v=2+2 \\\ & v=4m/s \\\ \end{aligned}$$ **Hence the correct option is D.** **Note:** If we didn’t consider the two separate functions for a while integrating for just consider the positive function (t-2) as at time 4 seconds it is represented by a positive function then the answer will be wrong. Because here we are integrating from the time when a particle was at rest to the time when it undergoes acceleration. Even if we just consider the positive function from limit 2 to 4 then we also have to change the limit of velocity and we don’t know the velocity of time at 2 seconds.