A car starts from rest at a stop sign. It accelerates at 4.0 m/s 2 for 6.0 s, coasts for 2.0 s, and then slows down at a rate of 3.0 m/s 2 for the next stop sign. How far apart are the stop signs?

The stop signs are 216 m apart.

Explanation:

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The equation of position and velocity of the car are the following:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position of the car at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

v = velocity of the car at time t.

Let´s place the origin of the frame of reference at the first stop sign so that x0 = 0. Let's calculate the traveled distance during the 6.0 s of acceleration at 4.0 m/s².

x = x0 + v0 · t + 1/2 · a · t² (x0 and v0 = 0)

x = 1/2 · a · t²

x = 1/2 · 4.0 m/s² · (6.0 s)²

x = 72 m

During the first 6.0 s, the car traveled 72 m.

Let's find the velocity reached by the car in that time:

v = v0 + a · t

v = a · t

v = 4.0 m/s · 6.0 m/s²

v = 24 m/s

After the first 6.0 s, the car travels for 2.0 s at v = 24 m/s and a = 0 m/s². The initial position in this case is the distance traveled so far, 72 m. So, after this 2.0 s the position of the car will be:

x = x0 + v · t

x = 72 m + 24 m/s · 2.0 s

x = 120 m

Then the car slows down until it stops at the next stop sign. Let's calculate the time it takes the car to stop, i.e. to reach a velocity of zero:

v = v0 + a · t

When the car stops, the velocity is zero:

0 = 24 m/s - 3.0 m/s² · t (notice that the acceleration is negative because the car is slowing down)

-24 m/s / -3.0 m/s² = t

t = 8.0 s

Then, the car travels 8.0 s slowing down. Let's calculate the position of the car after that time:

x = x0 + v0 · t + 1/2 · a · t²

The initial position and velocity, in this case, are the position and velocity of the car after the previous two parts of the travel:

x = 120 m + 24 m/s · 8.0 s - 1/2 · 3.0 m/s² · (8.0 s)²

x = 216 m

The stop signs are 216 m apart.

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