Elementary physics in 1 dimension

Click For Summary
SUMMARY

The discussion focuses on a physics problem involving a car and a truck, where the car accelerates to pass the truck traveling at a constant speed of 19.0 m/s. The car starts at the same speed and accelerates at 0.550 m/s², needing to cover a total distance of 50.4 m (the combined lengths of the car and truck plus the initial gap). The incorrect time of 13.5 seconds is derived from misunderstanding the point at which the car is considered to have passed the truck, which is when the rear of the car is 25.4 m ahead of the truck's front. The correct interpretation of the problem clarifies that the passing maneuver is complete when the car's front is ahead of the truck.

PREREQUISITES
  • Understanding of kinematic equations in one dimension
  • Familiarity with concepts of constant acceleration
  • Knowledge of relative motion between two objects
  • Ability to solve quadratic equations
NEXT STEPS
  • Study kinematic equations for uniformly accelerated motion
  • Learn about relative velocity and its applications in physics
  • Practice solving physics problems involving multiple objects in motion
  • Explore graphical representations of motion to visualize acceleration and distance
USEFUL FOR

Students studying physics, educators teaching kinematics, and anyone interested in understanding motion dynamics in one dimension.

parm12
Messages
3
Reaction score
0
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 23.7 m behind the truck's rear bumper. The car begins accelerating at a constant acceleration 0.550 m/s^2, then pulls back into the truck's lane when the rear of the car is a distance 25.4 m ahead of the front of the truck. The car is of length 4.90 m and the truck is of length 21.8 m. How much time is required for the car to pass the truck?

I'm having trouble understanding why 13.5 s is incorrect.

Here is my approach:

T_position = 19(m/s)t + 50.4m (50.4 = Car length + Truck Length + Distance in beteen)
C_position = .275(m/s^2)t^2 + 19(m/s)t

C_position corresponds to the rear of the car, and T_position the front of the truck.

Solving for both position equations yields:

50.4m = .275(m/s^2)t^2

sqrt(50.4/.275) = t = 13.5s
 
Physics news on Phys.org
They're probably asking when the car pulls back into the same lane as the truck. Then the car will have completed his manouver to pass the car, not when the car's rear is at the truck's front.
It isn't specifically mentioned, but from the information given in the question, it's highly certain.
 
Correct

You were correct, but I do feel misled by the question. The time it takes a particle to pass another particle is at the time t when the distances are equivalent (assuming the same conditions for motion as in this problem). It is afterwards from this time that one particle will be "passed" the other.

Thanks much for the input :approve:
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 4 ·
Replies
4
Views
12K
  • · Replies 6 ·
Replies
6
Views
9K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 6 ·
Replies
6
Views
7K
  • · Replies 15 ·
Replies
15
Views
4K
  • · Replies 4 ·
Replies
4
Views
8K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
7K