Solving Kinematics Homework - Car & Truck Displacement & Speed

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SUMMARY

The discussion focuses on solving kinematics problems involving a car and a truck with different acceleration and velocity. The car accelerates at 4.0 m/s² and covers a displacement of 400 m in 10 seconds, while the truck, moving at a constant speed of 90.0 km/h (25.0 m/s), covers 250 m in the same time frame. The car passes the truck at a distance of 312.5 m, and the calculations reveal that the car's speed at that moment is 25.0 m/s. The participants also address the braking distances required for both vehicles when encountering a barrier 100 m away.

PREREQUISITES
  • Understanding of kinematic equations, specifically d = 0.5at²
  • Knowledge of unit conversions, particularly converting km/h to m/s
  • Familiarity with concepts of acceleration and constant velocity
  • Ability to apply the equations of motion for uniformly accelerated motion
NEXT STEPS
  • Review the kinematic equations for uniformly accelerated motion
  • Practice converting between different units of speed and distance
  • Learn about the implications of uniform acceleration in real-world scenarios
  • Explore braking distance calculations and factors affecting stopping time
USEFUL FOR

This discussion is beneficial for students studying physics, particularly those focusing on kinematics, as well as educators looking for practical examples of motion and acceleration concepts.

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Homework Statement


" A car starts from rest and accelerates @ 4.0m/s^2 [E]. A truck with a constant velocity of 90.0km/h [E] overtakes and passes the car.
a) how far beyond the starting point is the car after 10.0 s?
How far beyond the starting point is the truck after 10.0s?
c) The car passes the truck @ a distance of 312.5m beyond the starting point. How fast is the car traveling at this instant?
d) How long does the car take to pass the truck?
e) Both drivers suddenly see a barrier 100.0m away and hit their breaks at the same time. If they take 3.0s to stop and decelerate uniformly, will they stop in time?


Homework Equations





The Attempt at a Solution



a) I found the displacement to be 400m [E]
b) I converted the 90km/h to 25.0m/s [E] and found the displacement to be 250m [E]
c) here's where I got stuck. I used the uaual velocity formula to solve for the change in time, but I got 12.5s as t. This doesn't make sense since in part a) I found the displacement to be 400m [E] in 10 seconds.
d) For this, I wanted to state that since the car encounters the truck at 312.5m [E], and the truck is traveling 25.0m/s [E]. it took the car 12.5 seconds to get there (calculated in pt c) but I think my logic and calculations in the previous step are wrong.
e) for the truck, I used the v2^2 = v1^2 + (2a)(displacement) to solve for displacement. I got 37.7m but since the last one was wrong..well its all connected and I'm frustrated.

Any help would be greatly appreciated
 
Last edited:
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Alyce777 said:

Homework Statement


" A car starts from rest and accelerates @ 4.0m/s^2 [E]. A truck with a constant velocity of 90.0km/h [E] overtakes and passes the car.
a) how far beyond the starting point is the car after 10.0 s?
How far beyond the starting point is the truck after 10.0s?
c) The car passes the truck @ a distance of 312.5m beyond the starting point. How fast is the car traveling at this instant?
d) How long does the car take to pass the truck?
e) Both drivers suddenly see a barrier 100.0m away and hit their breaks at the same time. If they take 3.0s to stop and decelerate uniformly, will they stop in time?


Homework Equations





The Attempt at a Solution



a) I found the displacement to be 400m [E]
b) I converted the 90km/h to 25.0m/s [E] and found the displacement to be 250m [E]
c) here's where I got stuck. I used the uaual velocity formula to solve for the change in time, but I got 12.5s as t. This doesn't make sense since in part a) I found the displacement to be 400m [E] in 10 seconds.
d) For this, I wanted to state that since the car encounters the truck at 312.5m [E], and the truck is traveling 25.0m/s [E]. it took the car 12.5 seconds to get there (calculated in pt c) but I think my logic and calculations in the previous step are wrong.
e) for the truck, I used the v2^2 = v1^2 + (2a)(displacement) to solve for displacement. I got 37.7m but since the last one was wrong..well its all connected and I'm frustrated.

Any help would be greatly appreciated

for a)

using
d=0.5at^2
you would get d=0.5*4*100=200m

You forgot to multiply it by 0.5.
 
rootX said:
for a)

using
d=0.5at^2
you would get d=0.5*4*100=200m

You forgot to multiply it by 0.5.

ooh thanks v. much
 

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