3 (a-c) (10%) Prolben 3: A 050-kg car comes to rest from a speed of 92.5 kmh in

Question

3 (a-c)

(10%) Prolben 3: A 050-kg car comes to rest from a speed of 92.5 kmh in a distance of 120 m. Assume the car is initially msoving in the positive direction 33% Part (a} f the brakes are tse on y thing nakine the car come to a stop. calculate be free in new ties direction of motion of the carj that the brakes apply ca the cr ea coeponent aleng the Grade Suemary Fb Dsductions Putatal i00% cotan asiacs atan) acotan)sinb) per atlenpt detailed viem +111213 Degrees Radians Submit lint Feedback: 1% dadaetuel peffondack 33% Part (b) Suppose instead of braking that the car bits a coccrete abutment al tell speed and is brought to a stop in 200 m Calelate the force, in newtons, exerted on the car in this case 3396 Part (c) what is the ratio ofthe force on the car fren the coacrete to the braking force?

Explanation / Answer

Here,

m = 950 Kg

final speed , v = 0 m/s

initial speed, u = 92.5 km/hr = 25.7 m/s

d = 120 m

a) let the acceleration is a

Using third equation of motion

v^2 - u^2 = 2 * a * d

0 - 25.7^2 = 2 * a * 120

a = -2.75 m/s^2

force acting on the car = m * a

force acting on the car = 2.75 * 950

force acting on the car = 2613 N

b) for d = 2 m

Using third equation of motion

v^2 - u^2 = 2 * a * d

0 - 25.7^2 = 2 * a * 2

a = 165 m/s^2

force acting on the car = m * a

force acting on the car = 165 * 950

force acting on the car = 1.57 *10^5 N

c)

ratio of forces = 1.57 *10^5/(2613)

ratio of forces = 60 times

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