Question

A small cart is rolling freely on an inclined ramp with a constant acceleration of .50 m/s2 in the x-direction. At time t=0, the cart has a velocity of 2.0 m/s in the +x-direction. If the cart never leaves the ramp, describe the motion of the cart at time t>5 s.

Answer 1

Explanation:

The given data is as follows.

             a = 0.5 $m/s^{2}$,     initial velocity = 2 m/s

After sometime, the inclined velocity will be equal to 0. Now, using the equation of motion as follows.

            $V^{o}_{f} - V_{o}$ = at

        t = $\frac{-V_{o}}{-a}$

          = $\frac{2 m/s}{0.5 m/s^{2}}$

          = 4 s

And, at t = 4 s,  $v_{f}$ = 0 and the cart starts to roll down the incline.

So, for t > 5 sec we assume that t = 6 sec.

Hence,   $v_{f} - v_{o} = at$

         $v_{f} = 0.5 m/s^{2} \times 6 sec$

         $v_{f}$ = 3 m/s

This means that the cart is travelling in -x direction and it is speeding at t > 5 sec.

Answer 2

Answer:

At t = 5 s, the velocity is 4.5 m/s, and for t > 5 s, it will continue increasing 0.5 m/s each second

Explanation:

We can find the velocity of the cart after t = 5 seconds using the equation:

v = u +at

where

v is the final velocity

u is the initial velocity

t is the time

a is the acceleration

For the cart in the problem,

$a=0.50 m/s^2\\u = 2.0 m/s$

Substituting t = 5 s, we find the velocity after 5 seconds:

$v=2.0+(0.50)(5)=4.5 m/s$

And after t > 5 s, the cart will continue accelerating, increasing its velocity by 0.50 m/s each second.