9) The next two questions refer to the situation below. At t=0, a green car is m

Question

9)The next two questions refer to the situation below.

At t=0, a green car is moving to the right with speed vo,g and is slowing down with an acceleration of magnitude a. Also at t=0, a red car, located a distance D to the right of the green car, is moving to the left with speed vo,r and is speeding up with an acceleration of magnitude a. The initial speed of the green car (vo,g) is twice the initial speed of the red car (vo,r). We define the x-direction to be positive to the right with x=0 at the position of the green car at t=0. Recall that speeds (vo,g and vo,r) are the magnitudes of the velocities and are thus always positive.

What is tx, the time that the two cars pass each other, in terms of vo,g, a, and D, as needed?

10) What is vrgvr(tx)vg(tx), the difference of the velocities of the two cars at the time they pass each other in terms of vo,g, a, and D, as needed?

-vrg=32vo,g+43aDvo,g

Explanation / Answer

Green Car -
Initial Speed = vo,g
Acceleration = -a

Red car -

Initial Speed = vo,g
Acceleration = a

vo,g = 2 * vo,r

Let they meet at time tx
Distance travelled by Green car in time tx =
S = ut + 0.5 at^2
s1 = vo,g * tx - 0.5 a tx^2

Distance travelled by Red car in time tx =
S = ut + 0.5 at^2
s2 = vo,r * tx + 0.5 a tx^2

Now s1 + s2 = D
vo,g * tx - 0.5 a tx^2 + vo,r * tx + 0.5 a tx^2 = D
vo,g * tx + vo,r * tx = D
vo,g * tx + vo,g /2 * tx =D
tx = 2D/3 * vo,g

(b)
vrg vr(tx)vg(tx)
vg(tx) = vo,g - a*tx
vg(tx) = vo,g - a* (2D/3 * vo,g) i

vr(tx) = - (vo,g /2 + a * (2D/3 * vo,g)) i

vrg = - vo,g /2 - a * (2D/3 * vo,g) - vo,g + a* (2D/3 * vo,g)
vrg = - vo,g * 3/2
vrg = -3/2 * vo,g

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