The driver of a car wishes to pass a truck that is traveling at a constant speed

Question

The driver of a car wishes to pass a truck that is traveling at a constant speed of 19.0 m/s. Initially, the car is also traveling at a speed 19.0 m/s and its front bumper is a distance 23.1 m behind the truck's rear bumper. The car begins accelerating at a constant acceleration 0.570 m/s^2, then pulls back into the truck's lane when the rear of the car is a distance 25.0 m ahead of the front of the truck. The car is of length 5.00 m and the truck is of length 21.3 m.

How much time is required for the car to pass the truck?

What distance does the car travel during this time?

What is the final speed of the car?

Explanation / Answer

Part A - How much time is required for the car to pass the truck? We need to find the total distance the car must travel so that the car’s rear is 25 m ahead of the front of the truck. Initially, the car’s front is 23.1 m behind the truck’s rear. If the truck was not moving, the car would have to travel 23.1 m so that the front of the car would be at the rear of the truck; another 21.3 m for the length of the truck, then another 25 m to put the front of the car 25 m ahead of the truck, then another 5 m for the length of the car to put the rear of the car 25 m ahead of the truck. 23.1 + 21.3 + 25 + 5 = 74.4 m, if the truck were not moving. We also have to take into account the velocity and acceleration of the vehicles. Initial position of the front of the truck in relation to the car’s rear bumper is 23.1 + 21.3 + 5 = 49.4 m Truck moves at a constant speed, so its acceleration is 0. Position equation for the truck: xtruck(t) = 49.4 m + (19.0 m/s) * t + (1/2) (0) t2 Position equation for the car: xcar(t) = 0 m + (19.0 m/s) * t + (1/2) (0.570 m/s2) * t2 We want to find the time when the car’s position equation is 25 m ahead of the truck’s or Xtruck(t) – xcar(t) = 25 m 25 m = 49.4 m + (19.0 m/s) * t – ((19.0 m/s) * t + (1/2) (0.570 m/s2) * t2) Solve for t: t = 16.16 sec = 16.2 sec Part B - What distance does the car travel during this time? xcar(t) = 0 m + (19.0 m/s) * t + (1/2) (0.570 m/s2) * t2 = 19(16.2) + (1/2)(0.570)*16.22) = 382.6 m Part C - What is the final speed of the car? Vcar(t) = (19.0 m/s) + (0.570 m/s2) * t = 19.0 + (0.570)(16.2) = 28.2 m/s

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