Question

Suppose that a car performs a uniform acceleration of 0.42 m/s from rest to 30.0 km/h in the first stage of its motion (From point A to point B). Then it moves at constant speed for half a minute (From point B to point C). After that the car applies the breaks, stopping the vehicle in a uniform manner while the vehicle travels an additional 7.00m distance (From point C to point D). (a) How far did the car travel from the starting point? (b) How long was the car in motion? (c) What is the average speed of the car during the entire motion?

Answer 1

Answer:

a)Total distance = 399. 5 m

b)Total time =51.51 sec

c)Average speed = 7.75 m/s

Explanation:

For A to B:

$S=ut+\dfrac{1}{2}at^2$

$v^2=u^2+2as$

v= u + at

8.33 = 0.42 x t

t=19.83 sec

1 Km/h=0.27 m/s

30 Km/h=8.33 m/s

$8.33^2=2\times 0.42\times s$

s=82.6 m

For B to C

V= 8.33 m/s

s= V x t

s=8.33 x 30

s=249.9 m

For C to D

$S=ut-\dfrac{1}{2}at^2$

v= u - at

Final speed v=0

So

s=v x t/2

7= 8.33 x t/2

t=1.68 sec

Total distance = 82.6 + 249 .9 +7

Total distance = 399. 5 m

Total time = 19.83 + 30 + 1.68

Total time =51.51 sec

Average speed =Total distance/Total time

Average speed = 399.5/51.5

Average speed = 7.75 m/s