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Veronika [31]
3 years ago
8

When the driver presses the brake pedal, his car stops with an acceleration of - 8.7 m/s. How far will the car travel while comi

ng to a complete stop if its initial speed is 10 m/s? A. 10.2 m B. 13.1 m C. 9.5 m D. 5.7m
Physics
1 answer:
klemol [59]3 years ago
3 0

Answer:

D. 5.7 m

Explanation:

The acceleration of the car is -8.7 m/s² and the car has an initial speed (u) of 10 m/s. The car said to come to rest, that means that the final velocity (v) of the car is 0 m/s.

To find the distance traveled by the car (s) before complete stop, this equation is being used:

v² = u² + 2as

2as=v^2-u^2\\s=\frac{v^2-u^2}{2a}\\ Substituting:\\s=\frac{0-10^2}{2*-8.7}\\ s=\frac{-100}{17.4}\\ s=5.7\ m

The car traveled 5.7 m before coming to a complete stop

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A car traveling at 15 m/s starts to decelerate steadily. It comes to a complete stop in 10 seconds. What is it’s acceleration
tresset_1 [31]

So the initial velocity is 15 m/s, the final velocity is 0 since it's at a complete stop and time is 10 seconds.  Therefore:

Acceleration=\frac{v_{final}-v_{initial}}{t_{final}-t_{initial}} =\frac{0m/s-15m/s}{10s-0s} =-1.5\frac{m}{s^2}

Therefore, the acceleration is -1.5 m/s^2.  The reason it's negative is due to the fact that the vector is going against it's original movement since it's decelerating.

7 0
2 years ago
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A rock is projected upward from the surface of the moon, at time t = 0.0 s, w a velocity of 30 m/s. The acceleration due to grav
Vinvika [58]
<h2>Answer: 277.777 m</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told that the rock was<u> projected upward from the surface</u>, we will only use the equations related to the Y axis.

In this sense, the movement equations in the Y axis are:

y-y_{o}=V_{o}.t+\frac{1}{2}g.t^{2}    (1)

V=V_{o}-g.t    (2)

Where:

y  is the rock's final position

y_{o}=0  is the rock's initial position

V_{o}=30\frac{m}{s} is the rock's initial velocity

V is the final velocity

t is the time the parabolic movement lasts

g=1.62\frac{m}{s^{2}}  is the acceleration due to gravity at the surface of the moon

As we know y_{o}=0 , equation (2) is rewritten as:

y=V_{o}.t+\frac{1}{2}g.t^{2}    (3)

On the other hand, the maximum height  is accomplished when V=0:

V=V_{o}-g.t=0    (4)

V_{o}-g.t=0    

V_{o}=g.t    (5)

Finding t:

t=\frac{V_{o}}{g}    (6)

Substituting (6) in (3):

y=V_{o}(\frac{V_{o}}{g})+\frac{1}{2}g(\frac{V_{o}}{g})^{2}    (7)

y_{max}=\frac{{V_{o}}^{2}}{2g}    (8)  Now we can calculate the maximum height of the rock

y_{max}=\frac{{(30m/s)}^{2}}{(2)(1.62m/s^{2})}   (9)

Finally:

y_{max}=277.777m  

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1. Why is it important to use units in any graph?
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Answer:

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Explanation:

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Answer:

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Explanation:

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Answer:a

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