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 20.5 . Initially, the car is also traveling at a speed 20.5 and its front bumper is a distance 23.6 behind the truck's rear bumper. The car begins accelerating at a constant acceleration 0.620 , then pulls back into the truck's lane when the rear of the car is a distance 25.5 ahead of the front of the truck. The car is of length 4.10 and the truck is of length 20.0 .
A - How much time is required for the car to pass the truck?
B - What distance does the car travel during this time?
C - What is the final speed of the car?

Explanation / Answer

(a) What you need to recognize to solve this problem is that the displacement of the car, although larger than the truck’s displacement, happens in the same time interval as the truck. The displacement of the car is the distance it is behind the truck (24.0m) plus the truck’s length (21.0m) plus the distance it pulls ahead of the truck (26.0m) or 71.0m. I don’t see the need for the car’s length because the initial lead of the truck is measured from the car’s FRONT bumper, and its lead over the truck at the end is measured from the car’s REAR bumper. This effectively negates the car’s length. The equation for the car’s total displacement then, is: ?x + 71.0m = (20.0m/s)t + 0.5(0.600m/s²)t²----------------->(1) The truck’s displacement (during which it has constant velocity) is: ?x = (20.0m/s)t ---------------------------->(2) Subtracting (2) from (1) eliminates ?x and allows you to find t: ?x + 71.0m - ?x = 0.5(0.600m/s²)t² 71.0m = (0.300m/s²)t² t = v[71.0m / 0.300m/s²] = 15.4s (b) The total distance the car travels through during this interval is: ?x = (20.0m/s)(15.4s) + 0.5(0.600m/s²)(15.4s)² = 379m (c) The car’s final speed is: v = v0 + at = 20.0m/s + (0.600m/s²)(15.4s) = 29.2m/s

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