Calculating Time from Acceleration and Length - Karl's Schoolbook Question

[kalle437]

Hello there!
I am a swedish student (16 years old) and I would need help with a part of a question in my book. I have a test tomorrow, and it would be nice if I could get some help with this. The thing I need to know is, is it possible to calculate time by only knowing acceleration and lenght?

The question was about a car breaking. I know the acceleration is -7,856 m/s^2, and the length is 22m.

I've been trying a lot but havn't found a way to calculate time with this information. The question wanted to know the velocity before the car started to break, but i figured I need time before I can calculate the velocity.

Ask if it was something you didn't understand, and thanks in advance for all the help! :)

Karl

 

[Doc Al]

You could find the time first, but you don't have to. It depends on what kinematic equations you have at your disposal. (You can find a list here: Basic Equations of 1-D Kinematics )

What's the most basic relationship between speed, acceleration, and time?

Hint: Try to solve it symbolically before plugging in any values.

 

[kalle437]

You could find the time first, but you don't have to. It depends on what kinematic equations you have at your disposal. (You can find a list here: Basic Equations of 1-D Kinematics )

What's the most basic relationship between speed, acceleration, and time?

Hint: Try to solve it symbolically before plugging in any values.

Forgot to mention my english ain't the best, so have understanding if I would misunderstand anything :)

The most basic relationship must be;

a = deltav / deltat

but, i don't know the starting velocity, only the ending velocity, which means I can't calculate time through this formula with this info, or can I?

 

[Doc Al]

That's a good start. No, you can't calculate the time directly, but you can express the time in terms of other variables. Since the final speed is zero, we can write: a = v/t or t = v/a. (Where v is the initial velocity, which is what we really want to find.)

How can you combine that with the distance that you're given? What's the most basic relationship between distance, time, and velocity?

 

[kalle437]

That's a good start. No, you can't calculate the time directly, but you can express the time in terms of other variables. Since the final speed is zero, we can write: a = v/t or t = v/a. (Where v is the initial velocity, which is what we really want to find.)

How can you combine that with the distance that you're given? What's the most basic relationship between distance, time, and velocity?

v=s/t

maybe we could replace v with s/t in the quation

t = v/a
t = (s/t)/a
ta = s/t
t^2*a = s
t^2 = s / a

t^2 = 22 / 7,865 ( the acceleration is negative, but since you can't "root" any negative numbers i put it as a positive number, is this right or wrong? )
t^2 = 2,78
t = 1,67s

 

[Doc Al]

v=s/t

Almost. But the speed is not constant, so you need the average velocity = s/t. How can you write the average velocity in terms of the initial and final velocities?

 

[kalle437]

Almost. But the speed is not constant, so you need the average velocity = s/t. How can you write the average velocity in terms of the initial and final velocities?

That would be ( Vzero + V ) / 2 :

Vzero = initial velocity
V = the final velocity

t = v/a
t = ( ( Vzero + V ) / 2 ) / a
ta = ( Vzero + V ) / 2
2ta = Vzero + V
Vzero = 2ta - V

BUT, this is 2 unknown variables ( Vzero and t ), which makes the equation uncalculable (as far as I know). Maybe its possible to replace t with something else that we do know, but the two equations that has been into discussion both include v.

 

[Doc Al]

That would be ( Vzero + V ) / 2 :

Vzero = initial velocity
V = the final velocity

Good. But I thought the final speed was zero and the initial speed is V. In any case, you can now write:
v/2 = s/t

Combine this with:

t = v/a

Eliminate t, since you don't really need it.

 

[kalle437]

maybe we could replace the Vzero with something instead.

V = Vzero + at
-Vzero = -V + at
Vzero = V - at

Vzero = 2ta - V
V - at = 2ta - V
V = 3at - V
2V = 3at

Since V = 0 the equation would be

0 = 3at
0 = 3 * 7,856 * t
0 = 23,56t

Didn't work. :/

EDIT : Was written before your post. Will check into it now.

 

[kalle437]

v/2 = s/t

You must mean Vzero + V = s / t ?
if we replace the t with v/a we get a deltav, and the delta v should be the negative of Vzero.

Because out of that I got:

( Vzero + V ) / 2 = s / ( -Vzero / a )
Vzero + V = 2s / (v / a )
-Vzero/a ( Vzero + V ) = 2s
-Vzero (Vzero+V) = 2as
-Vzero^2 - Vzero * V = 2as

Since V equals zero, we can delete that term (is it called term in english ?).

-Vzero^2 = 2as
-Vzero^2 = 345,7
-Vzero = 18,6 m/s
Vzero = -18,6 m/s

The answer is right, only thing is its negative. Maybe I put a negative number somewhere in the equation when I shouldn't have.

 

[malty]

-Vzero^2 = 2as
-Vzero^2 = 345,7
-Vzero = 18,6 m/s
Vzero = -18,6 m/s

The answer is right, only thing is its negative. Maybe I put a negative number somewhere in the equation when I shouldn't have.

How is this number +ve when car is decelerating??

 

[kalle437]

How is this number +ve when car is deccelerating??

Thanks :) Now the equation works, awesome! :D

Thanks a lot Doc Al for helping me through this, it means a lot :)