Question

A man is moving with 36 kmph. The time of reaction is 0.9 seconds. on seeing an obtacle in the path, he applies breaks and decelenrates at 5 m/s², the total distance covered before he stops ?​

Answer 1

Answer:

Answer is 19 m distance covered by man before he stops.

Explanation:

Given :

u = 36 km/h = 36x (5/18) = 10 m/s

Time of reaction, t = 0.9 s

decelration, a = -5 m/s^2

v = 0 m/s

Let s1 be the distance covered by man when the object is seen by him

we have, u = s1/t

because s1 = u x t

= 10 x 0.9

= 9 m

when he applies brakes and decelerates at the rate of 5 m/s^2, the distance covered by him is s2

We have, v^2 = u^2 + 2as2

therefore 0 = u^2 - 2as2 ( therefore v=0 m/s)

therefore s2 = u^2/2a = 10^2/ (2 x 5)

therefore s2 =10 m

So, total distance covered s = s1 + s2 = 9 +10 =19 m

Hence 19 m distance covered by man before he stops.

Answer 2

Man moves with 36 km/hr .

• His speed in m/s unit is 36 × 5/18 = 10 m/s.

His reaction time is 0.9 seconds, so distance travelled before braking is :

= 10 × 0.9 = 9 metres.

Then he retarded at 5 m/s².

Applying EQUATIONS OF KINEMATICS

${v}^{2} = {u}^{2} + 2as$

$\implies \: {0}^{2} = {10}^{2} + 2( - 5)s$

$\implies \: 10s = 100$

$\implies \: s = 10 \: metres$

So, total distance travelled is 9 + 10 = 19 metres.