Car Passing Truck: Solving for Time Required

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Homework Help Overview

The problem involves a car attempting to pass a truck while both vehicles are in motion. The car starts at a speed equal to that of the truck and accelerates to overtake it. The scenario includes specific distances and speeds, requiring the calculation of the time taken for the car to complete the maneuver.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the equations of motion for both the car and the truck, with one participant attempting to equate their positions to find the time required. Others suggest treating the problem from the truck's frame of reference to simplify calculations.

Discussion Status

There is an ongoing exploration of different methods to approach the problem, with participants questioning the setup and initial conditions. Some guidance has been offered regarding the choice of reference frame, but no consensus on the correct solution has been reached.

Contextual Notes

Participants are working with specific initial conditions and distances that may influence their calculations. The problem constraints include the lengths of the vehicles and their initial speeds, which are critical to the setup.

Toranc3
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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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