i need help with problem 2.66 (a) How much time does it take the car to overtake

Question

i need help with problem 2.66 (a) How much time does it take the car to overtake the truck? (b) How far was the car behind the truck initially? (c) What is the speed of each when they are abreast? (d) On a single graph, sketch the position of each vehicle as a function of time. Take x = 0 at the initial location of the truck. You are standing at rest at a bus stop. A bus moving at a constant speed of 5.00 m/s passes you. When the rear of the bus is 12.0 m past you, you realize that it is your bus, so you start to run toward it with a constant acceleration of 0.960 m/s^2. How far would you have to run before you catch up with the rear of the bus, and how fast must you be running then? Would an average college student be physically able to accomplish this? The driver of a car wishes to pass a truck that is traveling at a constant speed of 20.0 m/s (about 45 mi/h). Initially, the car is also traveling at 20.0 m/s, and its front bumper is 24.0 m behind the truck's rear bumper. The car accelerates at a constant 0.600 m/s^2, then pulls back into the truck's lane when the

Explanation / Answer

We will catch the bus that means distance covered by the bus and by us are equal, so

12+Vb*t = 1/2*a*t^2
12+5t = 0.48t^2
t = (5±(5^2-1.92*12))/0.96 = 6.67 sec

S = 12+Vb*t = 12+5*6.67 = 45.35 m
S = 1/2*a*t^2 = 0.48*6.67^2 = 21.35 m

Final speed is
Vfin = a*t = 0.96*6.67 = 6.4 m/sec (23.04 km/h)

For average college student this speed is more.

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