Question

A car is travelling along a road.

It passes point A at a constant
speed of V m/s, and drives for T seconds at this speed.

It then accelerates at a constant rate for 6 seconds until it reaches a speed
of 2V m/s. Maintaining this speed, it arrives at point B after a
further 2T seconds.

The total distance travelled between A and B
was 528 m and the average speed was 20 m/s.

Find V and T.

Answer 1

V = 12.279 m/s

T = 6.8 s

Step-by-step explanation:

First find the distance traveled during the acceleration.

Given:

u = V m/s

v = 2V m/s

t = 6 s

Find: s

s = ½ (v + u) t

s = ½ (2V + V) (6)

s = 9V

So the total distance is:

528 = (V)(T) + 9V + (2V)(2T)

528 = 5VT + 9V

The total time is:

T + 6 + 2T

3T + 6

Average speed = distance / time

20 = 528 / (3T + 6)

3T + 6 = 26.4

3T = 20.4

T = 6.8

Plugging in and solving for V:

528 = 5V (6.8) + 9V

528 = 43V

V = 12.279