Stopping distance

It points in a direction opposite to the action of the car. Given the mass of the car is 500 keg. Braking Force = mass x acceleration acceleration direction of motion -500 keg(-8 m s-2) = -4000 N braking force positive The braking force is in fact the friction between the trees and the road surface. When the brakes are applied, the wheels are locked and the trees rubs against the road surface. Skid mark is left on the road surface. The braking force decreases if the road surface is wet, the surfaces of the trees are worn out. Suppose the initial velocity of the car is 40 m s-l .

The reaction time of the driver remains as 0. 5 s and the braking force remains as 4000 N. For the same braking force and same mass, the acceleration is the same. The slope of the graph remains as -8 m s-2 parallel Will the car hit the man A with an initial separation of 40 m ? Stopping distance = shaded area = mom Stopping distance > mom . Therefore the car will hit the man A. Suppose the reaction time of the driver is 1 s. The initial velocity remains as 20 m s-l and the braking force remains as 4000 N. Will the car hit the man A with an initial separation of 40 m ?

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