A car is traveling on a flat, straight highway at a speed of 105km/hr. The drive

Question

A car is traveling on a flat, straight highway at a speed of 105km/hr. The driver sees cattle crossing the road 150m ahead and must stop. The car's brakes can de-accelerate the car at a rate of 2.25m/s^2.

Find: a) assuming the cow doesn't move, show how to determine whether the car is able to prevent a car-cow accident

b) If the car can't stop in time, determine the speed with which the car hits the cow

c) If there is an accident, the car will be forced to stop during a 1-second time interval. Show how to calculate the collision distance and acceleration that results from a collision.

Explanation / Answer

here,

initial speed of car , u = 105 km/h= 29.2 m/s

a)

s =150 m

a = - 2.25 m/s^2

let the stopping distance be s'

v^2 - u^2 = 2 * a * s'

- 29.2^2 = - 2 * 1.15 * s'

s' = 370 m

as s' > s

the car will hit the Cows

b)

the final speed of car , v^2 = u^2 + 2 * a * s

v = sqrt(29.2^2 - 2 * 150 * 2.25)

v = 13.3 m/s

c)

for the collison to take place

the initial speed , u = 13.3 m/s

t = 1 s

let the accelration be a'

v = u + a' * t

0 = 13.3 + a' * 1

a' = - 13.3 m/s^2

the collison distance , s = v^2 /2a

s = 13.3^2 /( 2 * 13.3) m

s = 6.65 m

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