Question

A car is moving on a straight road at a speed of 20 ms^-1 . The coefficient of friction between the tyres of the car and the road is 0.4 . Find the shortest distance within which the car can be stopped . Explain step by step and give correct answer ......​

Answer 1

Answer:

50 meter

Explanation:

U= 20 m/s

V=0 m/s

let g=10m/s²

Now, we have to find acceleration ( negative )

A= -0.4×10 = -4 m/s²

distance can be find by Newton's third equation of motion .

v²= u² + 2as

Then S,

0²=20²+2×-4× s

0= 400-8s

S= 400÷8

S= 50 m

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