Physics: 7. a roller coaster’s velocity at the top of a hill is 10 m/s. two seconds later it reaches the bottom ofthe hill with a velocity of 26 m/s. what was the acceleration of the coaster? 8. a roller coaster is moving at 25 m/s at the bottom of a hill. three seconds later it reaches the top ofthe hill moving at 10 m/s. what was the acceleration of the coaster? 9. 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? 10. a child drops

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7. a roller coaster’s velocity at the top of a

UnusualKc8513

05.01.2022

7) a=8m/s^2

8) a=-5m/s^2

9) a=-1.5m/s^2

10) v=29.4m/s

Explanation:

For the problems 7, 8 and 9 we just apply the definition of acceleration, since no more information is given, which is:

$a=\frac{\Delta v}{\Delta t}=\frac{v_f-v_i}{t_f-t_i}$

So for each problem we will have:

7) $a=\frac{26m/s-10m/s}{2s}=8m/s^2$

8) $a=\frac{10m/s-25m/s}{3s}=-5m/s^2$

9) $a=\frac{0m/s-15m/s}{10s}=-1.5m/s^2$

For the problem 10, we use the equation of velocity in accelerated motion:

v=v_0+at

Since the ball starts from rest and the acceleration is that of gravity (we take the downward direction positive), we have:

v=(0m/s)+(9.8m/s)(3s)=29.4m/s

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