A rock is dropped from a bridge. What happens to the magnitude of the acceleration and the speed of the rock as it falls? (neglect friction]
8. A brick is dropped from rest from a height of 4.9 m. How long does it take the brick to reach the ground?
9. The velocity of a car changes over an 8 s time period as shown in the following table.
(a) Plot the velocity-time graph of the motion.
(b) What is the displacement of the car during the entire 8 s?
(c) Find the slope of the line between t=0 s and 1=4 s? What does this slope represent?
(d) Find the slope of the line between t=5s and t=7s. What does this slope represent?
Pusition(m) 0 , 4 ,8 ,12 ,16, 20, 20 ,20
Time (0, 1, 2, 3, 4, 5)
1. As the rock falls, the magnitude of the acceleration remains constant. The speed of the rock increases as it falls.
8. The time taken for the brick to reach the ground can be calculated using the equation:
h = (1/2)gt^2, where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
In this case, h = 4.9 m. Rearranging the equation, we get:
t = sqrt(2h/g) = sqrt(2 * 4.9 / 9.8) = sqrt(0.99) ≈ 0.994 seconds. Therefore, it takes approximately 0.994 seconds for the brick to reach the ground.
9. (a) The velocity-time graph of the motion can be plotted using the given values as follows:
Time (s) Velocity (m/s)
0 0
1 4
2 8
3 12
4 16
5 20
6 20
7 20
8 20
(b) The displacement of the car during the entire 8 seconds can be calculated by finding the area under the velocity-time graph.
In this case, the area is a trapezoid with base 8 seconds and height 20 m/s. Thus, the displacement is given by:
Displacement = (1/2) * (8 + 20) * 8 = 156 meters.
(c) The slope of the line between t = 0 s and t = 4 s is given by the change in velocity divided by the change in time.
Here, the change in velocity is 16 m/s (from 0 to 16 m/s) and the change in time is 4 s. Therefore, the slope is:
Slope = (16 m/s) / (4 s) = 4 m/s^2.
This slope represents the acceleration of the car during the time interval.
(d) The slope of the line between t = 5 s and t = 7 s is given by the change in velocity divided by the change in time.
Here, the change in velocity is 0 m/s (velocity remains constant at 20 m/s) and the change in time is 2 s. Therefore, the slope is:
Slope = (0 m/s) / (2 s) = 0 m/s^2.
This slope represents the time interval during which the car maintains a constant velocity.