Question

a car is moving with a speed 20m/s. after it applying break it come at rest 5second ? find and calculate stopping distance ​

Answer 1

Answer:

v=u+at

20=a5

a=4m/s²

v²=u²+2as

20×20=s8

s=400/8

s=25m

Answer 2

ANSWER:-

stopping distance=50m

explanation:-

v= 0m/s. (it comes to rest)

u=20m/s. (given)

t=5sec

a=v-u/t

=

$= \frac{0 - 20}{5} \\ = - 4 {ms}^{ - 2}$

now,

${v}^{2} = {u}^{2} + 2as \\ \: {0}^{2} = { - 20}^{2} + 2 \times - 4 \times s \\ - 400 = - 8s \\ \frac{ - 400}{ - 8 } = s \\ 50m = s$

therefore stopping distance is 50m