A car is moving on a straight\road at a speed of 20 m/s1. The coeffici

1.	A car is moving on a straight\road at a speed of 20 m/s1. 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 (g = 10 m/s2).
Answer» Answer: 50M Explanation: 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

A car is moving on a straight\road at a speed of 20 m/s1. 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 (g = 10 m/s2).

Answer»

Answer: 50M

Explanation:

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

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A car is moving on a straight\road at a speed of 20 m/s1. 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 (g = 10 m/s2).​

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A car is moving on a straight\road at a speed of 20 m/s1. 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 (g = 10 m/s2).