Question

A can running at speed to 36 km/h changes its speed
to 90 km/h in 20 minutes. Find the acceleration and the
distance travelled by the car
during this time.​

Answer 1

Answer:

a = 0.0125 m/s

distance = 21 km

Explanation:

V = 90 km/h => 25 m/s

u = 36 km/h => 10 m/s

t = 20 min => 1200 s

V = u+at

25 = 10 + 1200a

15 = 1200a

a = 0.0125 m/s

s = 10*1200 + 1/2 * 0.0125 * 1200*1200

s = 12000 + 9000

s = 21000 m

s = 21 km

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Answer 2

Answer:

u = 36 km/hr = 36 * (5/18) = 10 m/s

v = 72 km/hr = 20 m/s

Using Newton's first eqn. of motion :

v = u + at

20 = 10 + a(5)

Hence a = 2 m/s²

Now the car stops in 20 seconds :

Using the same equation :

v = 0

u = 20 m/s

t = 20 sec

a = ?

0 = 20 + a (20)

a = -1 m/s²

In 2nd case there is retardation of 1 m/s²