A car is moving on a straight road at a speed of 20 m/s . 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.
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Answered by
155
U = initial speed = 20 m / s
V = Final velocity = 0
We take g = 10 m/s²
We need to find the acceleration which will be negative given that the car is coming to rest.
A = - 0.4 × 10 = -4 m / s²
From the formula :
V² = U² + 2as
We can get S
0 = 20² - 2× 4 × S
0 = 400 - 8S
S = 400/8 = 50m
V = Final velocity = 0
We take g = 10 m/s²
We need to find the acceleration which will be negative given that the car is coming to rest.
A = - 0.4 × 10 = -4 m / s²
From the formula :
V² = U² + 2as
We can get S
0 = 20² - 2× 4 × S
0 = 400 - 8S
S = 400/8 = 50m
Answered by
18
The answer is 50 m.
I hope solution is understood
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