How Fast Was the Car Going Before Braking?

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
7 replies · 2K views
r26h
Messages
9
Reaction score
0

Homework Statement


A car comes to a stop in 7.5 meters while braking at a rate of 2.0m/s each second.
A) What was the initial speed of the car?
B) What is the speed of the car after 0.5 seconds?

Homework Equations


Final velocity = Initial velocity + acceleration*time

The Attempt at a Solution


A) 0=Initial velocity+(-2)(7.5)
Initial velocty = 15 m/s

B)0=15+(-2)(.5)
Velocity=14m/s
 
Last edited by a moderator:
Physics news on Phys.org
r26h said:

Homework Statement


A car comes to a stop in 7.5 meters while breaking at a rate of 2.0m/s each second.
Braking at a rate of 2.0 m/s each second sounds like a deceleration.
A) What was the initial speed of the car?
B) What is the speed of the car after 0.5 seconds?

Homework Equations


Final velocity = Initial velocity + acceleration*time

The Attempt at a Solution


A) 0=Initial velocity+(-2)(7.5)
Initial velocty = 15 m/s
Did you check the units of this calculation? I don't think 2.0 m/s per second multiplied by 7.5 meters will give an answer in seconds
B)0=15+(-2)(.5)
Velocity=14m/s

A "breaking" car is on which goes to pieces. A "braking" car is one which comes to a stop using its brakes.
 
Ok, so what if to find the time I did 7.5m/ 2s/m = 3.75 seconds.

Then used that equation:
0=Initial velocity + (-2m/s)(3.75s)
So the initial speed is 7.5m/s? Does that make sense?
 
r26h said:
Ok, so what if to find the time I did 7.5m/ 2s/m = 3.75 seconds.
I guess you mean 7.5m/(2m/s), but you must not go plugging values into equations just because you know an equation that wants inputs of that type. You have to think about whether the equation is appropriate to the circumstance. What standard equation did you invoke to arrive at that calculation? What do the variables in the equation represent?
 
Ok, I see my error. What if I used Vf^2=Vi^2+2ax, rearranged to Vi=Square root of (Vf-2ax)
=square root (0-2(-2)(7.5)
initial velocity=square root of 30 or about 5.48 meters/second
 
To find the time, next I used Vf=Vi+at
0=5.48m/s +(-2m/s)(t)
-5.48m/s==2t
t=2.74s

Then which equation would I use to find the speed at 0.5 seconds?
 
r26h said:
To find the time, next I used Vf=Vi+at
0=5.48m/s +(-2m/s)(t)
-5.48m/s==2t
t=2.74s

Then which equation would I use to find the speed at 0.5 seconds?
You are not asked for the time to stop, are you? But the equation you used will also tell you the speed after a given time.