A car is initially at rest on a straight road. The histogram below shows the car

Question

A car is initially at rest on a straight road. The histogram below shows the car's acceleration along that road as a function of time. Calculate the speed of the car at t = 4 s. 7.00 m/s Calculate the distance traveled during the first 6 s. 42.0m Calculate the distance traveled from t = 10 s to t = 13 s. 30.0 m Calculate the car's average speed from t = 6 s to t = 9 s. 1.33m/s The average speed is the distance traveled divided by the time elapsed. Try using this graph to create a velocity vs. time graph, where the slopes of the new graph are given by the accelerations on this graph.

Explanation / Answer

0 to 1 sec:

a = 0

v = 0

d = 0

1 to 3 sec:

a = 3 m/s^2

v0 = 0

d1 = v0t + a t^2 /2 = (0 x 2) + (3 x 2^2 / 2) = 6 m

3 to 5 sec:

a = 1 m/s^2

v0 = 0 + (3 x 2) = 6m/s

t = 2 sec

d2 = (6 x 2) + (1 x 2^2 / 2) = 14 m

5 to 6 sec :

a = -1 m/s^2

v0 = 6 + (2 x 1) = 8 m/s

d3 = (8 x 1) + (-1 x 1^2 / 2) = 7.5 m/s

d = d1 + d2 + d3 = 27.5 m .............Ans (distance travelled during first 6 sec)

average speed = distance travelled / time taken

6 to 7 sec:

v0 = 8 + (-1 x 1) = 7 m/s

a = 3 m/s^2

d4 = (7 x 1) + (3 x 1^2 / 2) = 8.5 m

7 to 8sc:

v0 = 7 + (3 x 1) = 10 m/s

a= 2 m/s^2

d5 = (10) + (2 x 1^2 / 2) = 11 m

8 to 9 sec:

v0 = 10 + (2 x 1) = 12 m/s

a = -1 m/s^2

d6 = (12 x 1) - (1 x 1^2 / 2) = 11.5 m

distance = 31 m/s

time = 9 -6 = 3 sec

average speed = 31 / 3 = 10.33 m/s ..............Ans

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