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A speed ramp at an airport is basically a large conveyor belt on which you can stand and be moved along the belt of one ramp moves at a constant speed such that a person who stands still on it leaves the ramp 64 s after getting on .Clifford is in a real hurry however and skips the speed ramp starting from rest with an acceleration of 0.37 m/ s2 he covers the same distance as the ramp does but in one-fourth the time . What is the speed at which the belt of the ramp is moving
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Guest User
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Sahil Khan
1

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}

Qalaxia Master Bot
0

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