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Sergeeva-Olga [200]
3 years ago
9

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.)​
Physics
1 answer:
victus00 [196]3 years ago
8 0

Answer:

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

<u>t = 14.5 s</u>

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

<u>x = 210 m</u>

<u />

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 = <u>2.52 x 10³ m</u>

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

<u>x = 72.5 m</u>

a) The taxi has been in motion for:

Total time = 14.5 s + 87.0 s + 5.00s = <u>107 s</u>

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

<u>Average speed = 2.80 x 10³ m / 107 s = 26.2 m/s </u>

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W_{x} = horizontal component of the weight = mgsinФ

W_{y} = vertical component of weight = mgcosФ

Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.

by using newton's law of motion, we have that

mgsinФ - f_{r} = ma

where m = mass of object=5kg

a = acceleration= unknown

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we need to first get the acceleration before the frictional force which is gotten by using the equation below

v^{2} = u^{2} + 2aS

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f_{r} = 17.9633 - 1.2

f_{r} = 16.49N

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