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

The velocity of the cart at time t=5 s is 4.5 m/s. After 5 seconds, the cart will continue to accelerate down the ramp with a constant acceleration of 0.5 m/s2 in the x-direction, assuming it does not leave the ramp.

Step-by-step explanation:

Given the information that a small cart is rolling freely on an inclined ramp with a constant acceleration of 0.5 m/s2 in the x-direction and an initial velocity of 2.0 m/s in the +x-direction at time t=0, we can describe the motion of the cart at time t>5 s. The cart's velocity at any time t can be calculated using the equation v = v0 + at, where v is the velocity at time t, v0 is the initial velocity, a is the acceleration, and t is the time.

Using this equation, the velocity of the cart at time t=5 s would be:

v = 2.0 m/s + (0.5 m/s2 × 5 s) = 2.0 m/s + 2.5 m/s = 4.5 m/s

Since the cart never leaves the ramp and there are no forces acting in the horizontal direction other than gravity and the resultant acceleration, the motion of the cart after t>5 s would involve it continuing to accelerate down the ramp at a constant acceleration of 0.5 m/s2.

Answer 2

Step-by-step 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 =
`(-V_(o))/(-a)`

=
`(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) * 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.