A zero-friction (air bearing) cart is heading downhill on a slightly inclined tr

Question

A zero-friction (air bearing) cart is heading downhill on a slightly inclined track. At a moment when the cart’s speed downhill is 2.15 m/s, a fan thruster on the cart is activated. The cart immediately begins to experience constant acceleration, such that it slows in its motion, stops for an instant (1.86 s after the fan is turned on), and begins to drive uphill.

Analyze the motion beginning when the fan is turned on.

a) How far does the cart travel from activation until the time it reaches a momentary stop?

b) How long will it take to get all the way back to where it started?

c) What will its speed be at that time?

Explanation / Answer

Let us assume cart moves from A(uphill) to a point B(downhill where it reaches speed 2.15 m/s) to a point C ( downhill where it stops and then turns back) to a point D( uphill where it started from)

speed at point A =va = 0

speed at point B = vb= 2.15 m/s

acceleration after fan is turned on = a = ?

time after which it stops after fan is turned on = t = 1.86 s

speed at point C where it stops for a moment = vc = 0 m/s

a) let us assume cart travels distance ‘d’ after activation of fan till the point where it stops for a moment

vc = vb +at

0 = 2.15 + a(1.86)

a = -1.16 m/s^2

Also, vc^2 = vb^2 + 2ad

0^2 = 2.15^2 + 2(-1.16)d

d = 1.99 m

b)To travel back to point D, it has to cover distance CD = CB + BA = 1.99 + BA

We need to calculate BA first

Since cart acquires speed 2.15 m/s starting from rest while travelling distance BA

Kinetic energy of cart = Work done by gravitational force

0.5 mv^2 = mg(BA)

0.5 x m x (2.15)^2 = m x 9.81 x (BA)      [ acceleration is taken as g as cart is slightly inclined]

(BA) = 0.236 m

CD = 1.99 +0.236 = 2.23 m

Now cart will travel uphill with initial velocity = 0m/s and acceleration = 1.16 m/s^2 (uphill) and it will cover distance CD = 2.23 m

Let time taken is t’

Using displacement-time equation

2.23 = 0 + 0.5(1.16)t’^2

t’ = 3.84 s

c) Speed at point D = vd

vd= vc +at’

vd = 0 +(1.16)(3.84) = 4.45 m/s

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