What Is the Correct Acceleration When an Object Stops Suddenly?

[Nat3]

Homework Statement

An object traveling at a constant 100km/hr (27.8 m/s) stops in 0.75m

Homework Equations

What was the object's acceleration?

The Attempt at a Solution

$a = \frac{\Delta v}{\Delta t}$

$\Delta t = \frac{0.75 m}{27.8 m/s} = 0.027 s$

$a = \frac{0-27.8 m/s}{0.027 - 0 s} = -1029.63 m/s^2$

The textbook answer is -500 m/s^2. I'm confused :(

 

[lewando]

$\Delta t = \frac{0.75 m}{27.8 m/s} = 0.027 s$

Your problem is with this line. It would be appropriate if velocity was constant over that distance, but it is not.

 

[gneill]

When you calculate your Δt, the velocity is not constant over the time interval. Thus $d \neq \Delta v\; \Delta t$. Rather, the average velocity will be Δv/2, so that
$$d = \frac{\Delta v}{2} \Delta t$$
$$\Delta t = \frac{2 d}{\Delta v}$$

 

[Nat3]

When you calculate your Δt, the velocity is not constant over the time interval. Thus $d \neq \Delta v\; \Delta t$. Rather, the average velocity will be Δv/2, so that
$$d = \frac{\Delta v}{2} \Delta t$$
$$\Delta t = \frac{2 d}{\Delta v}$$

Why is the average velocity Δv/2? What does "d" represent?

Thanks for your help!

 

[gneill]

Why is the average velocity Δv/2? What does "d" represent?

Thanks for your help!

d is the distance traveled (you can call it Δx if you wish). The initial velocity is v, the final velocity is zero. So the Δv is (v - 0). The average is (v + 0)/2, or Δv/2.

 

[Nat3]

The average is (v + 0)/2, or Δv/2.

Is this average really accurate? For example, if a car drives at 10 m/s for 10 minutes, then slows to a stop in one second, is the average speed really going to be (0+10)/2 ?

 

[gneill]

Is this average really accurate? For example, if a car drives at 10 m/s for 10 minutes, then slows to a stop in one second, is the average speed really going to be (0-10)/2 ?

It is for the period of the deceleration; the rest of the trip is not being considered

And it's (10 + 0)/2 in this case.

 

[Nat3]

It is for the period of the deceleration; the rest of the trip is not being considered

And it's (10 + 0)/2 in this case.

So this works because we assume the acceleration (deceleration) during the .75m the object is slowing down is constant? If the acceleration changed during that time, then your method of calculating the average velocity would not be accurate?

Thanks :)

 

[gneill]

So this works because we assume the acceleration (deceleration) during the .75m the object is slowing down is constant? If the acceleration changed during that time, then your method of calculating the average velocity would not be accurate?

Thanks :)

Correct.