Braking speed and distance traveled

[emily081715]

Homework Statement

The acceleration of a particular car during braking has magnitude bt, where t is the time in seconds from the instant the car begins braking, and b = 2.2 m/s3 .If the car has an initial speed of 60 m/s , how far does it travel before it stops?

Homework Equations

: [/B]
I know that basic kinematic equations will not work for this equation because acceleration is not constant. Therefore the equations have to be integrated from the given equation of acceleration

The Attempt at a Solution

:[/B]
when i integrated i got the equations:
a=2.2t

v= 2.2/2 t^2 +60

x= 2.2/6 t^3 +60t

I'm trying to figure out how to isolate for t in order to solve for distance, but when i set the equation for distance equal to zero i ended up getting the wrong answer while solving for t. if someone can point me in the right direction on how to solve for t, i know how to finish the problem solving for distance. i just am unsure about how to get time

 

[kuruman]

Hi Emily081715 and welcome to PF



Your expression for v says that v increases with time. Is that what happens to a car that is braking?
Also, x is not zero when the car stops. What is zero when that happens?

 

[emily081715]

You expression for v says that v increases with time. Is that what happens to a car that is braking?

should my equations say -60? If that was the case shouldn't my acceleration be negative as well since its braking and slowing down?
Also, if x isn't zero when the car stops, is the final velocity zero ? that would mean i need to set that equation equal to zero to solve for time

 

[kuruman]

When t is equal to zero, v = 60 m/s as it should. When t is greater that zero, v starts to decrease. Where should the negative sign go?

 

[emily081715]

When t is equal to zero, v = 60 m/s as it should. When t is greater that zero, v starts to decrease. Where should the negative sign go?

the negative sign goes infront of the 60. making the equation v= 2.2/2t^2- 60t

 

[kuruman]

But now when you set t = 0 in your expression, you get zero, not 60 m/s.

 

[emily081715]

But now when you set t = 0 in your expression, you get zero, not 60 m/s.

should it then be infront of 2.2

 

[David Lewis]

The magnitude of acceleration is positive even when the car is slowing down. The plus and minus sign give you the directional sense of the quantity. If we say the sense of the car's velocity is positive then the sense of the acceleration would be negative.

 

[kuruman]

should it then be infront of 2.2

Yes. Leave the 60 as you had it. Now can you find the time of travel? It is the same as the time it takes the car to stop.

 

[kuruman]

The magnitude of acceleration is positive even when the car is slowing down. The plus and minus sign give you the directional sense of the quantity. If we say the sense of the car's velocity is positive then the sense of the acceleration would be negative.

That's another of saying that (in 1-dimension), when the velocity and the acceleration have the same sign, the speed is increasing; when they have opposite signs, the speed is decreasing.

 

[emily081715]

That's another of saying that (in 1-dimension), when the velocity and the acceleration have the same sign, the speed is increasing; when they have opposite signs, the speed is decreasing.

Okay so i changed the sign of acceleration and used the equation v=-2.2/2 t^2 +60 to solve for t. i got (10 √66)/11 for time. when plugging that answer in for distance i got 295.4 meters. does this answer seem correct?

 

[kuruman]

It is correct.

 

[emily081715]

It is correct.

thanks for the help