Question

The distance required for a car to come to a stop will vary depending on how fast the car is moving. Suppose that a certain car traveling down the road at a speed of 10 m / s can come to a complete stop within a distance of 20 m . Assuming the road conditions remain the same, what would be the stopping distance required for the same car if it were moving at speeds of 5 m / s , 20 m / s , or 40 m / s ?

Answer 1

5 m

80 m

320 m

Step-by-step explanation:

`v_(o)` = Initial speed of the car = 10 ms⁻¹

`v_(f)` = Final speed of the car = 0 ms⁻¹

`d` = Stopping distance of the car = 20 m

`a` = acceleration of the car

On the basis of above data, we can use the kinematics equation

`v_(f)^(2) = v_(o)^(2) + 2 a d\\0^(2) = 10^(2) + 2 (20) a\\a = - 2.5 ms^(-2)`

`v_(o)` = Initial speed of the car = 5 ms⁻¹

`v_(f)` = Final speed of the car = 0 ms⁻¹

`d'` = Stopping distance of the car

`a` = acceleration of the car = - 2.5 ms⁻²

On the basis of above data, we can use the kinematics equation

`v_(f)^(2) = v_(o)^(2) + 2 a d'\\0^(2) = 5^(2) + 2 (- 2.5) d'\\d' = 5 m`

`v_(o)` = Initial speed of the car = 20 ms⁻¹

`v_(f)` = Final speed of the car = 0 ms⁻¹

`d''` = Stopping distance of the car

`a` = acceleration of the car = - 2.5 ms⁻²

On the basis of above data, we can use the kinematics equation

`v_(f)^(2) = v_(o)^(2) + 2 a d''\\0^(2) = 20^(2) + 2 (- 2.5) d''\\d'' = 80 m`

`v_(o)` = Initial speed of the car = 40 ms⁻¹

`v_(f)` = Final speed of the car = 0 ms⁻¹

`d'''` = Stopping distance of the car

`a` = acceleration of the car = - 2.5 ms⁻²

On the basis of above data, we can use the kinematics equation

`v_(f)^(2) = v_(o)^(2) + 2 a d'''\\0^(2) = 40^(2) + 2 (- 2.5) d'''\\d''' = 320 m`