A car starts from rest and travels for 5.0 s with a uniform acceleration of +2.9 m/s^2. The driver then applies the brakes, causing a uniform acceleration of −2.0 m/s^2. If the brakes are applied for 3.0 s, determine each of the following.
(a) How fast is the car going at the end of the braking period? ______ m/s
(b) How far has the car gone? _______ m
An explanation would be great! Thanks!
(a) How fast is the car going at the end of the braking period? ______ m/s
(b) How far has the car gone? _______ m
An explanation would be great! Thanks!
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I'll show you a graphical method, which is easier to visualize.You can start by drawing a simple velocity/time sketch graph, with velocity on the vertical axis and time on the horizontal one. If you mark the velocity axis in multiples of 2.9 m/s, then its easy to see what's happening.
V:- 0 2.9 5.8 8.7 11.6 14.5
t :- 0 1 2 3 4 5
so you can see that the velocity changes by 2.9 m/s after every second has passe. After 5 secs, its velocity is therefore 14.5 m/s
The brakes are then applied for 3 secs and the decelleration is -2.0 m/s. So
V:- 14.5 12.5 10.5 8.5
t :- 5 6 7 8
a) So after 8 secs its final velocity will be 8.5 m/s
b) Recall that the total distance travelled is the area under the velocity time graph, since, if the graph was a rectangle, the area would be V x t = (m/s) * s = m.
If you join a line from the final velocity point to the time axis, you form a 4-sided figure. There are a few ways to work out the area of this figure. I dropped a vertical line from the max velocity point to the time axis and a horizontal line from this to the final velocity point. That gives me two triangles and a recangle.
Area of large triangle = (1/2) * 5 * 14.5 = 36.25 m^2
Area of small triangle = (1/2) * 3 * (14.5 - 8.5) = 9 m^2
Area of rectangle = 3 * 8.5 = 25.5 m^2
Total distance travelled = 36.25 + 9 + 25.5 = 70.75 m
V:- 0 2.9 5.8 8.7 11.6 14.5
t :- 0 1 2 3 4 5
so you can see that the velocity changes by 2.9 m/s after every second has passe. After 5 secs, its velocity is therefore 14.5 m/s
The brakes are then applied for 3 secs and the decelleration is -2.0 m/s. So
V:- 14.5 12.5 10.5 8.5
t :- 5 6 7 8
a) So after 8 secs its final velocity will be 8.5 m/s
b) Recall that the total distance travelled is the area under the velocity time graph, since, if the graph was a rectangle, the area would be V x t = (m/s) * s = m.
If you join a line from the final velocity point to the time axis, you form a 4-sided figure. There are a few ways to work out the area of this figure. I dropped a vertical line from the max velocity point to the time axis and a horizontal line from this to the final velocity point. That gives me two triangles and a recangle.
Area of large triangle = (1/2) * 5 * 14.5 = 36.25 m^2
Area of small triangle = (1/2) * 3 * (14.5 - 8.5) = 9 m^2
Area of rectangle = 3 * 8.5 = 25.5 m^2
Total distance travelled = 36.25 + 9 + 25.5 = 70.75 m