A motorcycle is stopped by the side of the road when a car passes at 50 mi/h. Tw

Question

A motorcycle is stopped by the side of the road when a car passes at 50 mi/h. Twenty seconds later the motercycle starts chasing the car. Assume that the motorcycle accelerates at 8 ft/s^2 until it reaches 60 mi/h and then travels at a constant speed. Fine the ammount of time it will take the motorcycle to overtake the car and the total distance traveled by the motorcycle in that time.

-I keep getting weird answers & I know it can't be correct, please show me the work with how its done! please, i need to learn this

Explanation / Answer

50 mi/h = 73.3333 ft/s..........60 mi/h = 88 ft/s

Total distance travelled by both will be same

Now suppose :

Motorcycle acclerates for time t1 and moves with constant speed for t2

Total distance travelled by car = (73.333 ft/s)*(20 + t1+t2) = 1466.6667 + 73.333*(t1+t2)

Total distance travelled by motor cycle = 0.5*8*t1^2 + 88*t2 = 4*t1^2 + 88*t2

so equating distances :

1466.6667 + 73.333*(t1+t2) = 4*t1^2 + 88*t2

and we will have t1 from kinematic equation

v = u + a*t

88 = 0 + 8*t1

t1 = 11 s

so

1466.6667 + 73.333*(t1+t2) = 4*t1^2 + 88*t2

1466.6667 + 73.333*(11+t2) = 4*121 + 88*t2

(88-73.3333)*t2 = 1789.33323

t2 = 122 sec

Total time = 20+ t1 + t2 = 20+11 + 122 = 153 s

Total distance = 73.3333*153 = 11220 ft = 2.125 miles

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