What is the acceleration 𝑎y of the yellow car?

Question

A blue car and a yellow car start from rest at a red stop light. When the light turns green, both cars accelerate forward.

The blue car accelerates uniformly at a rate of 3.3 m/s2 for 3.0 s. It then continues at a constant speed for 6.8 s. Finally, the blue car brakes such that its speed decreases uniformly until it comes to rest 95.54 m from where it started.

The yellow car accelerates uniformly for the entire distance, finally catching the blue car just as the blue car comes to a stop.

Answer 1

BB,

I want to focus on just the blue car. With that information, you should be able to calculate the necessary acceleration of the yellow car. Let's track the blue car;s travel in three phases:

Phase 1) Initial positive acceleration from rest (let's assume to the right).

Start at x_s = 0, v₀ = 0 (started at rest), traveled for t = 3 s, at ∝_b1 = 3.3 m/s², and Phase 1 end at position x₁.

The formula for displacement d(t) = d(t) = 1/2∝_yt² + v₀t + d₀ allows us to calculate x1 because we know alpha, v0 and d0, so x1 = 1/2 (3.3)(3)² + 0(3) + 0 = 14.85 m

The formula for velocity v(t) = ∝(t) + v0 allows us to calculate velocity at the end of phase 1.

v(3) = 3(3.3) + 0 = 9.9 m/s

Phase 2) No acceleration (at a constant velocity after its initial acceleration)

Start at x₁, ∝_b2 = 0 (not accelerating), but traveling at some constant velocity v₂=9.9 m/s, traveling for t = 6.8 s, and Phase 2 ends at position x₂.

The displacement equation gives us x₂ = d(t) = 1/2(0)(6.8)² + 9.9(6.8) + 14.85 = 82.17 m

and the velocity is unchanged at the end of the phase at 9.9 m/s

Phase 3) Final negative acceleration (to the left) until it returned to rest.

Start at x₂, ∝_b3 is negative because it is slowing down, velocity begins at 9.9 m/s, but the final velocity = 0 (because it stops) at end of Phase 3 and the phase lasts some unknown amount of time t₃.

We need to get our final velocity to 0, so using the velocity equation:

v(t₃) = ∝(t₃) + 9.9 or 0 = ∝(t3) + 9.9 so ∝ = - 9.9/t₃

We also need to displace from 82.17 m to our final position of 95.54 m or 13.37 m

95.54 = 1/2∝t₃² + (9.9)t₃ + 82.17, and substituting the velocity equation into this displacement we get

95.54 - 82.17 = (1/2)(-9.9/t₃)t₃² + 9.9t₃

13.37 = -4.45t₃ + 9.9t₃ = 4.45 t₃

t₃ = 13.37/4.45 = 3.00 s (To 2 significant digits after the decimal point)

(we don't need to solve for the acceleration here but it is -9.9/3 = -3.3 m/s²)

Now you know the total displacement of the yellow car and the total time that displacement took, so you should be able to use the displacement equation to compute the constant acceleration. Send me a message if you need more help.

Answer 2

First step - find t, the time that the blue car is in motion.

Then we know that after t seconds, the yellow car has travelled 95.54 m, from this we can find the yellow cars acceleration from

s = ut + 1/2 a t^2 (u=0, s= 95.54) then we can solve for a.

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Finding t.

First 3 seconds of blue car motion.

s = ut + 0.5 a t^2

u = 0

a = 3.3

t = 3

s = 14.85 m

The speed then is v = u + at, v = 0 + 3 x 3.3 = 9.9 m/s

It continues at this speed for 6.8 s, travelling a distance of 9.9 x 6.8 = 67.32 m

So after 3 + 6.8 = 9.8s the blue car has travelled 14.85 + 67.32 = 82.17 m

It comes to a stop after 95.54 m, ie it decelerates uniformly over a distance of 95.54 - 82.17 = 13.37 m

v^2 = u^2 + 2as. v = 0, u = 9.9, s = 13.37 giving a = -3.6653 m/s/s

v = u + at. v = 0, u = 9.9, a = -3.6653 giving t = 2.7 s

So blue car is in motion for 3.3 + 6.8 + 2.7 = 12.8 s

Yellow car is also in motion for 12.8s, travelling 95.54 m from rest with uniform acceleration

s = ut + 0.5 a t^2

s = 95.54

u = 0

t = 12.8

giving a = 1.167 m/s/s