How Do You Calculate Deceleration from Average Velocity?

[Nirupt]

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

 

[tms]

Start by writing down the equations that describe the car's motion.

 

[Nirupt]

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

 

[SammyS]

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

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

 

[tms]

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.

 

[Nirupt]

Do it one step at a time.

You have, v_avg = (v₀ + v)/2 , and you know v_avg ≈ 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

 

[SammyS]

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 !

 

[hms.tech]

Looks good !

No, there still seems to be an arithmetic mistake

[itex]V_{0}[/itex][itex]\approx[/itex]27.8 m/s

 

[Nirupt]

I'm not sure why I typed that but I know the answer is correct.