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

Answer:

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:

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