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Answer :
1) First, let's calculate the value of deceleration a that the car can achieve, using the following relationship:
[tex]2aS = v_f^2-v_i^2 = -v_i^2[/tex]
where S=67 m is the distance covered, vf=0 is the final velocity of the car, and vi=15 m/s is the initial velocity. From this we can find a:
[tex]a= \frac{-v_i^2}{2S}= \frac{-(15m/s)^2}{2\cdot 67 m}=-1.68 m/s^2 [/tex]
2) Then, we can assume this is the value of acceleration that the car is able to reach. In fact, the force the brakes are able to apply is
[tex]F=ma[/tex]
This force will be constant, and since m is always the same, then a is the same even in the second situation.
3) Therefore, in the second situation we have a=-1.68 m/s^2. However, the initial velocity is different: vi=45 m/s. Using the same formula of point 1), we can calculate the distance covered by the car before stopping:
[tex]2aS=-v_i^2[/tex]
[tex]S= \frac{-v_i^2}{2a} = \frac{-(45 m/s)^2}{2\cdot (-1.68 m/s^2)}=603 m [/tex]
[tex]2aS = v_f^2-v_i^2 = -v_i^2[/tex]
where S=67 m is the distance covered, vf=0 is the final velocity of the car, and vi=15 m/s is the initial velocity. From this we can find a:
[tex]a= \frac{-v_i^2}{2S}= \frac{-(15m/s)^2}{2\cdot 67 m}=-1.68 m/s^2 [/tex]
2) Then, we can assume this is the value of acceleration that the car is able to reach. In fact, the force the brakes are able to apply is
[tex]F=ma[/tex]
This force will be constant, and since m is always the same, then a is the same even in the second situation.
3) Therefore, in the second situation we have a=-1.68 m/s^2. However, the initial velocity is different: vi=45 m/s. Using the same formula of point 1), we can calculate the distance covered by the car before stopping:
[tex]2aS=-v_i^2[/tex]
[tex]S= \frac{-v_i^2}{2a} = \frac{-(45 m/s)^2}{2\cdot (-1.68 m/s^2)}=603 m [/tex]
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