A car, shown in Figure P1.44, weighing 3000 lb tows a single axle two-wheel trai

Question

A car, shown in Figure P1.44, weighing 3000 lb tows a single axle two-wheel trailer

weighing 1500 lb at 60 mph. There are no brakes on the trailer, and the car, which

by itself can decelerate at 0.7g, produces the entire braking force. Determine the

force applied to slow the car and trailer. Determine the deceleration of the car and

the attached trailer. How far do the car and trailer travel in slowing to a stop? How

many seconds does it take to stop?

A car, shown in Figure P1.44, weighing 3000 lb tows a single axle two-wheel trailer weighing 1500 lb at 60 mph. There are no brakes on the trailer, and the car, which by itself can decelerate at 0.7g, produces the entire braking force. Determine the force applied to slow the car and trailer. Determine the deceleration of the car and the attached trailer. How far do the car and trailer travel in slowing to a stop? How many seconds does it take to stop?

Explanation / Answer

speed = 60 mph = 88 ft/sec

retardation = 0.7 g = 0.7 * 32.2 = 22.54 ft/sec2

total mass = 3000 + 1500 = 4500 lb

given car's self retardation = 0.7 g = 22.54 ft/sec2

so retardation force by car = mass * aceelration = 22.54 * 3000 = 67620

now force will be same ...

so rertardation of car with the traileer = force / total mass = 67620 / 4500 = 15.0267 lb ft/sec2 = 15.0267/32.2 g = 0.467 g

b) initial speed (u) = 60 mph = 88 ft/sec

retardation (a)= -15.0267 lb ft/sec2

final speed (v) = 0

v^2 = u^2 + 2 * a * s

so.. 0 = 88^2 - 2 * 15.0267* s

so.. distance travelled = s = 257.675 ft

c) v = u + a * t

0 = 88 - 15.0267 * t

so time = t = 5.856 secs

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