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29 November, 09:51

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?

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  1. 29 November, 11:03
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    The stop signs are 216 m apart.

    Explanation:

    Hi there!

    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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