Car Passing Truck: Solving for Time Required

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The discussion revolves around calculating the time required for a car to pass a truck while both are initially traveling at the same speed of 20 m/s. The car accelerates at 0.600 m/s² and must cover a total distance that includes its own length, the distance behind the truck, the truck's length, and the distance ahead needed to safely merge back. The initial calculations led to a time of 12.8 seconds, but the correct answer is 15.9 seconds, indicating a miscalculation in determining the total distance to be covered. A suggestion was made to analyze the problem from the truck's frame of reference, simplifying the calculations by treating the truck as stationary. This approach allows for easier computation of the time required for the car to complete the maneuver.
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Homework Statement



The driver of a car wishes to pass a truck that is traveling at a constant speed of 20m/s. Initially, the car is also traveling at 20m/s and its front bumper is 24m behind the trucks rear bumper. the car accelerates at a constant 0.600 m/s^2 , then pulls back into the trucks lane when the rear of the car is 26m ahead of the trucks front. the car is 4.5m long and the truck is 21m long.
a.) how much time is required for the car to pass the truck?

Homework Equations



x=xo+vo*t+1/2*a*t^(2)

The Attempt at a Solution



Car:

xc=xo+vo*t+1/2*a*t^(2)
xc=20m/s*t + 0.300m/s^(2)*t^(2)

Truck:

xt=xo+vo*t+1/2*a*t^(2)
xt=49.5m+20m/s*t

xt=xc

49.5m+20m/s*t=20m/s*t + 0.300m/s^(2)*t^(2)

I get t=12.8 seconds but the answer is 15.9seconds. What did I do wrong?
I equated the equations to find out when they are both at the same position.
 
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Here's how I would do it:

Treat the rear of the car as x0. The total distance it has to cover is its own length, plus the 24m, plus 21m (length of truck) plus the additional 26m before it pulls into the truck's lane.

Now initially they're traveling at the same constant velocity so you can treat it as if they're both initially at rest. So then all you have to solve is:

x_f=\frac{1}{2}at^2
75.5=0.3t^2

Which should give the desired answer.
 
Toranc3 said:
xt=49.5m+20m/s*t
Where did the 49.5 come from?

You'll find it a little easier to work in the truck frame, so you can ignore it's motion, and set the car's initial speed to zero.
 
tms said:
Where did the 49.5 come from?

You'll find it a little easier to work in the truck frame, so you can ignore it's motion, and set the car's initial speed to zero.

I set my origin at the rear of the car, which is how I got 49.5m for the truck. Can you explain more on working the problem in the trucks frame? Thanks!
 
Toranc3 said:
I set my origin at the rear of the car, which is how I got 49.5m for the truck. Can you explain more on working the problem in the trucks frame? Thanks!
Choose the frame in which the truck is at rest. Since it is traveling at a constant speed, it stays at rest in that frame, so it can basically be ignored. Since the car and truck start at the same speed, it also starts at rest in that frame.

Using such a frame just makes the calculations a bit easier.
 
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