A motorcycle starts from rest at and travels along a straight road with a constant acceleration of 2 6 / ft s until it reaches a speed of 50 / ft s. Afterwards it maintains this speed. Also, when t s  0, a car located 6000 ft down the road is traveling toward the motorcycle at a constant speed of 30 / ft s. Determine the time and the distance traveled by the motorcycle when they pass each other.

Time = 77.63 s

Distance = 3673.3 ft

Explanation:

Using equation of motion

v = u + at'

50 = 6 * t'

t' = 50 / 6

t' = 8.33 s

v² = u² + 2a (s - s•)

50² = 0² + 2 * 6 * (s - 0)

2500 = 12 * s

s' = 2500 / 12

s' = 208.3 ft

At t' = 8.33 the distance traveled by the car is

s'' = v• * t'

s'' = 30 * 8.33

s'' = 250 ft

Now, the distance between the motorcycle and car is

6000 - 208.33 - 250 = 5541.67

For motorcycle, when passing occurs,

s = v• * t, x = 50 * t''

For the car, when passing occurs,

s = v• * t, 5541.67 - x = 30 * t''

Solving both simultaneously, we have

5541.67 - [50 * t''] = 30 * t''

5541.67 = 30t'' + 50t''

5541.67 = 80t''

t'' = 5541.67 / 80

t'' = 69.3 s, substituting t'' = 69.3 back, we have

x = 50 * 69.3

x = 3465 ft

Therefore for motorcycle,

t = 69.3 + 8.33

t = 77.63 s

S = 3465 + 208.3

s = 3673.3 ft