Question

A cab driver picks up a customer and delivers her 2.03 km away, driving a straight route. The driver accelerates to the speed limit and, upon reaching it, begins to decelerate immediately. The magnitude of the deceleration is 5.65 times the magnitude of the acceleration. Find the lengths of the (a) acceleration and (b) deceleration phases of the trip.

Answer 1

during acceleration

x1 = vox*t + (1/2)*a1*t^2

x1 = 0 + (1/2)*a1*t1^2

speed after t1 time

v1x = vox + a1*t1

v1x = a1*t1

during decelaration

x2 = (v2x^2 - v1x^2)/(2*a2)

x2 = (0-(a1*t1)^2)/(2*-a2)

x2 = (a1*t1)^2/(2*a2)

x1 + x2 = 2.03 km = 2030 m

2030 = (1/2)*a1*t1^2 + (a1*t1)^2/(2*a2)

a2 = 5.65*a1

2030 = (1/2)*a1*t1^2 + a1^2*t1^2/(2*5.65*a1)

2030 = (1/2)*a1*t1^2 + 0.0885*a1*t1^2

a1*t1^2 = 3450

length of phase during acceleration x1 = (1/2)*a1*t1^2 = 1725 m = 1.725 km

part(b)

length of path during deceleration x2 = 2030 - 1725 = 305 m = 0.305 km