Calculating Time for Side-by-Side Cars: Car Kinematics Homework

In summary, the red car will have traveled 3.84 meters in 8.92 seconds, while the blue car will have traveled 21.5 meters in 8.92 seconds.
  • #1
Youngstabullz
2
0

Homework Statement



A red car is sitting at a stop light at rest. When the light turns green two things happen simultaneously; (1) the red car accelerates to a speed of 34.3 m/s in 8.92 seconds and then maintains that 34.3 m/s speed and (2) a blue car passes by the red car at a constant speed of 21.5 m/s.

Red Car: Vo=0 Vf=34.3 m/s a=3.84 (34.3/8.92)
Blue Car: Vo=21.5 m/s Vf=21.5 m/s

How much time (in seconds) will have elapsed from when the light turned green until the two cars are side-by-side?

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Homework Equations



distance = 1/2(Vo + Vf)t
distance = Vot + 1/2at^2


The Attempt at a Solution



I thought I would make them equal to each other, 1/2(Vo +Vf)t=Vot +1/2at^2 And find out what t equals to find out the time however It has yet to work after many attempts. Vo Any help is appreciated.
 
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  • #2
Welcome to PF!

Youngstabullz said:
A red car is sitting at a stop light at rest. When the light turns green two things happen simultaneously; (1) the red car accelerates to a speed of 34.3 m/s in 8.92 seconds and then maintains that 34.3 m/s speed and (2) a blue car passes by the red car at a constant speed of 21.5 m/s.

Red Car: Vo=0 Vf=34.3 m/s a=3.84 (34.3/8.92)
Blue Car: Vo=21.5 m/s Vf=21.5 m/s

How much time (in seconds) will have elapsed from when the light turned green until the two cars are side-by-side?

I thought I would make them equal to each other, 1/2(Vo +Vf)t=Vot +1/2at^2 And find out what t equals to find out the time however It has yet to work after many attempts.

Hi Youngstabullz! Welcome to PF! :smile:

Show us your calculations, and then we can see where you've gone wrong. :smile:

(you have allowed for the period of constant speed of 34.3, haven't you?)
 
  • #3
x(blue)=21.5*8.92+21.5*t

x(red)=.5*34.3*8.92+34.3*t

and so t=3.03 then the answer is answer=3.03+8.92=11.95 seconds
 
  • #4
rado5 said:
x(blue)=21.5*8.92+21.5*t

x(red)=.5*34.3*8.92+34.3*t

and so t=3.03 then the answer is answer=3.03+8.92=11.95 seconds

Hi rado5! :smile:

You must stop giving full answers to other people's questions (you've done it before).

You haven't even given Youngstabullz the chance to reply. :frown:

On this forum, we try just to give helpful hints, so that the OP can work the problem out themselves.

Full answers should only be given when all hints have failed.

Please restrain yourself. :smile:

(And your answer was wrong … again! :rolleyes:)
 
  • #5
Thankfully, I got the answer before you guys had responded to my question. :D
 
  • #6
Youngstabullz said:
Thankfully, I got the answer before you guys had responded to my question. :D

That's the idea! :smile:
 

What is car kinematics?

Car kinematics is the study of the motion of cars, including their position, velocity, and acceleration, without considering the forces that cause the motion.

What are the key variables in car kinematics?

The key variables in car kinematics are displacement, velocity, and acceleration. Displacement is the change in position, velocity is the rate of change of displacement, and acceleration is the rate of change of velocity.

What is the difference between speed and velocity in car kinematics?

Speed is a scalar quantity that only describes the magnitude of an object's motion. Velocity, on the other hand, is a vector quantity that describes both the magnitude and direction of an object's motion.

What is the equation for calculating acceleration in car kinematics?

The equation for calculating acceleration in car kinematics is a = (vf - vi)/t, where a is the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time interval.

How does car kinematics relate to other branches of physics?

Car kinematics is closely related to other branches of physics, such as mechanics, dynamics, and kinematics. It uses principles from these branches to study the motion of cars and how they interact with their environment.

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