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.)

Question

Physics

Gabriela Bell

3 January, 01:31

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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Answers (1)

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Isla · Answer 1

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