Question

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 a ball from a window. The ball strikes the ground in 3.0 seconds. What is thevelocity of the ball the instant before it hits the ground?

Answer 1

7)
`a=8m/s^2`

8)
`a=-5m/s^2`

9)
`a=-1.5m/s^2`

10)
`v=29.4m/s`

Step-by-step explanation:

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

`a=(\Delta v)/(\Delta t)=(v_f-v_i)/(t_f-t_i)`

So for each problem we will have:

7)
`a=(26m/s-10m/s)/(2s)=8m/s^2`

8)
`a=(10m/s-25m/s)/(3s)=-5m/s^2`

9)
`a=(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`