A ball starts at the origin from rest rolling down a hill. It takes seven seconds for the ball to reach point “A”, which is 25 meters from the origin. At point “B”, the ball’s velocity is 15 m/s. You may assume constant acceleration.
a. What is the acceleration of the ball? [2]
b. What is the velocity of the ball at point “A”? [2]
c. What is the distance between points “A” and “B”? [2]
d. SKETCH the corresponding velocity-time graph for the ball between the origin and point “B” [1]
a. OA = 0.5a*t^2 = 25 m.
0.5a*7^2 = 25
0.5a*49 = 25
a = 1.02 m/s^2
b. V = a*t = 1.02 * 7 = 7.14 m/s.
c. AB = (V^2-Vo^2)/2a
AB = (15^2-7.14^2)/2.04 = 85.3 m.
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To answer these questions, we can use the equations of motion. Let's start with question a:
a. What is the acceleration of the ball?
To find the acceleration, we can use the equation:
v = u + at
where:
- v is the final velocity
- u is the initial velocity
- a is the acceleration
- t is the time taken
In this case, the ball starts from rest, so the initial velocity (u) is 0 m/s. The final velocity (v) at point B is given as 15 m/s. The time taken (t) is the time taken to reach point A, which is 7 seconds.
Plugging the values into the equation, we can solve for the acceleration (a):
15 = 0 + a * 7
Simplifying the equation, we have:
15 = 7a
Dividing both sides by 7, we get:
a = 15/7
Therefore, the acceleration of the ball is approximately 2.143 m/s^2.
b. What is the velocity of the ball at point "A"?
To find the velocity at point A, we can use the equation:
v = u + at
Again, the initial velocity (u) is 0 m/s, the acceleration (a) is 15/7 m/s^2, and the time taken (t) is 7 seconds.
Plugging the values into the equation, we can solve for the final velocity (v):
v = 0 + (15/7) * 7
Simplifying the equation, we have:
v = 15
Therefore, the velocity of the ball at point A is 15 m/s.
c. What is the distance between points "A" and "B"?
To find the distance between two points, we can use the equation:
s = ut + (1/2)at²
where:
- s is the distance traveled
- u is the initial velocity
- a is the acceleration
- t is the time taken
In this case, the initial velocity (u) is 0 m/s, the acceleration (a) is 15/7 m/s^2, and the time taken (t) is 7 seconds.
Plugging the values into the equation, we can solve for the distance (s):
s = 0 * 7 + (1/2) * (15/7) * (7^2)
Simplifying the equation, we have:
s = 0 + (1/2) * 15 * 7
s = 52.5
Therefore, the distance between points A and B is 52.5 meters.
d. SKETCH the corresponding velocity-time graph for the ball between the origin and point "B"
To sketch the velocity-time graph, we can represent time on the x-axis and velocity on the y-axis.
Since the ball starts from rest at the origin, the initial velocity is 0 m/s. At point B, the velocity is given as 15 m/s. Therefore, we can draw a straight line connecting these two points, with time increasing along the x-axis and velocity increasing along the y-axis.
The graph will be a straight line starting from the origin and sloping upwards to a point at (7,15).