We use the kinematics equation

x = v_{0}t + \frac{1}{2}at^{2} to find the distance Clifford travels.

where v_{0} is the initial velocity, a is the acceleration, and t s the time.

It is given that the time he covers is \frac{1}{4}^{th} of the actual time.

So this is \frac{1}{4}\cdot 64 s = 16 s.

His initial velocity is zero, and he has an acceleration of 0.37 \frac{m}{s^{2}}

Inputting these values we get

x = 0 \times 16 s + \frac{1}{2} \times 0.37 \frac{m}{s^{2}} \times (16 s)^{2}

x = 47.36 m

It is given that he covers the same distance as the ramp. So the length of the ramp is also 47.36 m.

We have to calculate the speed at which the belt of the ramp is moving.

Speed is nothing but velocity with direction.

The speed = distance/time = \frac{47.76 m}{64 s} = 0.74 \frac{m}{s}

I found an answer from www.numerade.com

SOLVED:**Kinematics** in **One Dimension** | **Physics 2012** | Numerade

**Kinematics** in **One Dimension**, **Physics 2012** - John D. Cutnell, ... by an angry bear is running in a **straight line** toward his car at a **speed** of 4.0 m/s .

For more information, see SOLVED:**Kinematics** in **One Dimension** | **Physics 2012** | Numerade