Question

a car accelerat uniformly from 18km/h to 36km/ h in 5 sec. calculate the acceleration and distance covered by the car​

Answer 1

Explanation:

a = (v-u)/t

u = 18 km/h = 18000/3600m/s = 6m/s

v = 36 km/h = 36000/3600m/s = 10m/s

t = 5s

a = (10-6)/5 = 4/5m/s^2

2as = v^2 - u^2

2*4s/5 = 100 - 36

8s/5 = 64

s = (64*5)/8

s = 8*5 = 40m

Therefore acceleration is 4/5m/s^2 and distance is 40m.

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Answer 2

$\huge {\mathfrak {\underline {\underline {\pink {Answer}}}}}$

Explanation:

a = (v-u)/t

u = 18 km/h = 18000/3600m/s = 6m/s

v = 36 km/h = 36000/3600m/s = 10m/s

t = 5s

a = (10-6)/5 = 4/5m/s^2

2as = v^2 - u^2

2*4s/5 = 100 - 36

8s/5 = 64

s = (64*5)/8

s = 8*5 = 40m

Therefore acceleration is 4/5m/s^2 and distance is 40m.