Question

A cart is released from rest at the top of a long ramp at time $t = 0$ seconds, and moves down the ramp at a constant acceleration rate. At a time of $t$ , the cart has reached a speed of $2m{s^{ - 1}}$ . How fast will the cart be moving at the time of $3t$ ?
$(A)3m{s^{ - 1}}$
$(B)6m{s^{ - 1}}$
$(C)12m{s^{ - 1}}$
$(D)18m{s^{ - 1}}$

Answer 1

Hint : To solve this question, we are going to first see the initial and the final velocities and then, using the first equation of motion, we deduce an equation. After that, we take the condition at time $3t$ , and again using the first equation of motion and using the above derived equation, velocity is found.
The first equation of motion is
$v = u + at$

Complete Step By Step Answer:
As it is given here, that the cart is released from rest, this says that
Now, after time $t$ , the speed of the cart,
$v = 2m{s^{ - 1}}$
Using the first equation of motion, we get,
$v = u + at$
Putting the values of the initial and final velocities,
2 = 0 + at \\\ \Rightarrow at = 2 \\\
At the time $3t$ , let the velocity be $v'$ , now, initial velocity remains the same. So, using the equation $v = u + at$ , again, we get
\Rightarrow v' = 0 + a\left( {3t} \right) \\\ \Rightarrow v' = 0 + 3at \\\ \Rightarrow v' = 3\left( 2 \right) = 6m{s^{ - 1}} \\\

Note :
The two factors that affect the velocity of the cart at any point of the time are the initial velocity of the cart and the acceleration of the cart and the time at which we need to find the velocity. Also, Newton's first equation of motion depends on the uniformly accelerated motions i.e. with constant acceleration only.