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10 February, 13:31

A taxi traveling along a straight section of road starts from rest, accelerating at 2.00 m/s^2 until it reaches a speed of 29.0 m/s. Then the vehicle travels for 87.0 s at constant speed until the brakes are applied, stopping the vehicle in a uniform manner in an additional 5.00 s. How long is the taxi in motion (in s) ? What is the average velocity of the taxi for the motion described? (Enter the magnitude in m/s.)

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  1. 10 February, 14:34
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    a) The taxi is 107 s in motion

    b) The average velocity is 26.2 m/s

    Explanation:

    First, the car travels with an acceleration of 2.00 m/s². The equations for position and velocity that apply for the car are:

    x = x0 + v0 t + 1/2 a t²

    v = v0 + a t

    where

    x = position at time t

    x0 = initial position

    v0 = initial speed

    t = time

    a = acceleration

    v = speed

    Let's calculate how much distance and for how long the taxi travels until it reaches a speed of 29.0 m/s:

    Using the equation for velocity:

    v = v0 + a t

    v - v0 / a = t

    (29.0 m/s - 0 m/s) / 2 m/s² = t

    t = 14.5 s

    Then, in the equation for position:

    x = x0 + v0 t + 1/2 a t²

    x = 0 + 0 + 1/2 * 2.00 m/s² * (14.5 s) ²

    x = 210 m

    Then, the vehicle travels at constant speed for 87 s. The distance traveled will be:

    x = v * t

    x = 29.0 m/s * 87.0 s = 2.52 x 10³ m

    Lastly the car stops (v = 0) in 5 s. In this case, the car has a constant negative acceleration:

    Using the equation for velocity:

    v = v0 + a t

    if v=0 in 5 s, then:

    0 = 29.0 m/s + a * 5.00 s

    a = - 29.0 m/s / 5.00 s

    a = - 5.80 m/s²

    Using now the equation for the position, we can calculate how far has the taxi traveled until it came to stop:

    x = x0 + v0 t + 1/2 a t²

    x = 0 + 29.0 m/s * 5.00 s - 1/2 * 5.80 m/s² * (5.00s) ²

    x = 72.5 m

    a) The taxi has been in motion for:

    Total time = 14.5 s + 87.0 s + 5.00s = 107 s

    Note that we have always used x0 = 0, then, we have calculated the displacement for each part of the trip.

    Adding all the displacements, we will get the total displacement:

    Total displacement = 210 m + 2.52 x 10³ m + 72.5 m = 2.80 x 10³ m

    Average speed = total displacement / total time

    Average speed = 2.80 x 10³ m / 107 s = 26.2 m/s
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