How Do You Calculate Deceleration from Average Velocity?

In summary, the car's average velocity during deceleration was 50 km/h. Its acceleration was -2.37 m/s2 while slowing to a stop.
  • #1
Nirupt
35
0

Homework Statement

A car decelerates uniformly and comes to a stop after 10 s. The car's average velocity during deceleration was 50 km/h. What was the car's acceleration while slowing to a stop?

{Since the acceleration is UNIFORM (constant), use vavg = (v0 + v)/2... }

Homework Equations


The Attempt at a Solution



V0 is not known, Final V = 0, t = 10, Vavg is 13.9 m/s and trying to find a

But when I try to plug it in I feel I have to solve for v0
 
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  • #2
Start by writing down the equations that describe the car's motion.
 
  • #3
The equation I come up with was 13.9(10) = (v0 - 0/2)(10)
But that equals to -27.8 m/s^2 which is not the right answer
 
  • #4
Nirupt said:

Homework Statement



A car decelerates uniformly and comes to a stop after 10 s. The car's average velocity during deceleration was 50 km/h. What was the car's acceleration while slowing to a stop?

{Since the acceleration is UNIFORM (constant), use vavg = (v0 + v)/2... }

Homework Equations



The Attempt at a Solution



V0 is not known, Final V = 0, t = 10, Vavg is 13.9 m/s and trying to find a

But when I try to plug it in I feel I have to solve for v0
Do it one step at a time.

You have, vavg = (v0 + v)/2 , and you know vavg ≈ 13.9 m/s and v = 0 . So solve that for v0 .

After that it's pretty straight forward to find the acceleration.
 
  • #5
Nirupt said:
The equation I come up with was 13.9(10) = (v0 - 0/2)(10)
But that equals to -27.8 m/s^2 which is not the right answer
Write the equations using symbols, not numbers. It is easier to see what is happening with the physics that way.
 
  • #6
SammyS said:
Do it one step at a time.

You have, vavg = (v0 + v)/2 , and you know vavg ≈ 13.9 m/s and v = 0 . So solve that for v0 .

After that it's pretty straight forward to find the acceleration.

So 13.9m/s = (v0 + 0)/2 = 23.7 m/s
V0≈23.7 m/s
V≈0
t≈10
Then Δv/Δt..

0-23.7m/s
-----------
10 s

= -2.37 m/s2
 
  • #7
Nirupt said:
So 13.9m/s = (v0 + 0)/2 = 23.7 m/s
V0≈23.7 m/s
V≈0
t≈10
Then Δv/Δt..

0-23.7m/s
-----------
10 s

= -2.37 m/s2
Looks good !
 
  • #8
SammyS said:
Looks good !

No, there still seems to be an arithmetic mistake :-p

[itex]V_{0}[/itex][itex]\approx[/itex]27.8 m/s
 
  • #9
I'm not sure why I typed that but I know the answer is correct.
 

FAQ: How Do You Calculate Deceleration from Average Velocity?

What is Uniform Accelerated Motion?

Uniform Accelerated Motion is the motion of an object where the acceleration remains constant throughout its entire path. This means that the speed of the object increases or decreases by the same amount every second.

What is the difference between Uniform Accelerated Motion and Uniform Motion?

The main difference between Uniform Accelerated Motion and Uniform Motion is that in Uniform Motion, the speed of the object remains constant while in Uniform Accelerated Motion, the speed of the object changes at a constant rate.

What is the formula for calculating Uniform Accelerated Motion?

The formula for calculating Uniform Accelerated Motion is v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.

What is the difference between acceleration and deceleration?

Acceleration is the rate at which an object's speed or velocity changes over time, while deceleration is the rate at which an object's speed decreases over time. In Uniform Accelerated Motion, acceleration can be either positive or negative, depending on whether the object is speeding up or slowing down.

What are some real-life examples of Uniform Accelerated Motion?

Some real-life examples of Uniform Accelerated Motion include a car accelerating from 0 to 60 mph in a straight line, a roller coaster going down a steep hill, and a ball being thrown straight up in the air and falling back down due to the acceleration of gravity.

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