Question

7. A car travelling at 108 km/hour has its
speed reduced to 36 km/hour while tra-
velling a distance of 200m. Its acceleration
and the time taken are​

Answer 1

Answer:

ANSWER

V

2

=108×

18

5

=30m/s V

1

=36×

18

5

=10m/s

⇒V

2

=U

2

+2as

⇒(30)

2

=(30)

2

+2×a×200

⇒a=−2m/s

Answer 2

Answer:

Therefore, acceleration = -2 m/s^2 and time = 10 s

Explanation:

Given: Initial velocity (u) = 108 km/hour = 30 m/s

           Final velocity (v) = 36 km/hour = 10 m/s

We must convert to m/s because distance is in m.

           Distance = 200 m

To find: Acceleration and Time Taken

Solution: By equation of motion,

v^2 + u^2 = 2as

a = v^2 - u^2 / 2 s

a = 10^2 - 30^2 / 2 (200)

a = 100-900 / 400

a = -800 / 400 = -2 m/s^2

It is deceleration.

a = v-u/t

-2 = 10-30/t

-2 = -20/t

-2t = -20

t = -20/-2

t = 10 seconds.

Therefore, acceleration = -2 m/s^2 and time = 10 s