A ball is rolled up a ramp with an initial velocity of +10m/s. It arrives at the bottom of the ramp with the same speed 5 seconds later. What is the ball acceleration for the whole trip? What is the velocity after 2 seconds? After 3 seconds?
a = change in velocity/change in time
= (-10 - 10)/5 = -20/5 = -4 m/s^2
v = vi + a t
at t = 2
v = 10 -4*2 = 2 m/s
at t = 3
v = 10 -4*3 = -2 m/s
To find the acceleration of the ball for the whole trip, we can use the kinematic equation:
v = u + at
Where:
- v is the final velocity
- u is the initial velocity
- a is the acceleration
- t is the time
In this case, the initial velocity (u) is +10m/s, and the final velocity (v) is also +10m/s, since the ball arrives at the bottom of the ramp with the same speed. The time taken for the trip is 5 seconds.
Using the equation, we can rearrange it to solve for acceleration (a):
a = (v - u) / t
Substituting the given values into the equation, we have:
a = (+10m/s - +10m/s) / 5s
a = 0m/s / 5s
a = 0m/s²
Therefore, the ball has an acceleration of 0m/s² for the whole trip.
To find the velocity after 2 seconds and 3 seconds, we can use the same equation:
v = u + at
For 2 seconds:
v = +10m/s + 0m/s² * 2s
v = +10m/s
Therefore, the velocity of the ball after 2 seconds is +10m/s.
For 3 seconds:
v = +10m/s + 0m/s² * 3s
v = +10m/s
Similarly, the velocity of the ball after 3 seconds is also +10m/s.