How Long Does It Take for a Car to Stop with Friction After Engine Shutdown?

[Troubadour]

Homework Statement

If the engine of a car provides an acceleration of 2ms^-2 to start it from rest, assuming the mass to be roughly 1000kg, calculate the time after which the car comes to rest if the engine is turned off after 15 seconds and the frictional force is 15N)

Homework Equations

v=u + at
F= m*a

The Attempt at a Solution

v= 0 + (2)(15) = 30ms^-1
F= m*a = -15N= 1000 * a
Therefore, acc. caused by friction= (-15)/1000= -0.015
v=u+at, but car comes to rest, so v = 0
0= 30 +(-0.015)(t) = 2000sec

I'm not really sure of the answer as it seems to be quite impractical (:P).
Thanks.
~Troubadour

 

[spl-083902]

Looks like you did it right.

Other than writing
0= 30 +(-0.015)(t) = 2000sec
which is technically wrong since 0 != 2000sec. You should put t=2000s on a separate line

 

[Troubadour]

Looks like you did it right.

Other than writing
0= 30 +(-0.015)(t) = 2000sec
which is technically wrong since 0 != 2000sec. You should put t=2000s on a separate line

Oops...I didn't quite notice that...Thanks a lot for your help!

 

[PeroK]

Homework Statement

If the engine of a car provides an acceleration of 2ms^-2 to start it from rest, assuming the mass to be roughly 1000kg, calculate the time after which the car comes to rest if the engine is turned off after 15 seconds and the frictional force is 15N)

Homework Equations

v=u + at
F= m*a

The Attempt at a Solution

v= 0 + (2)(15) = 30ms^-1
F= m*a = -15N= 1000 * a
Therefore, acc. caused by friction= (-15)/1000= -0.015
v=u+at, but car comes to rest, so v = 0
0= 30 +(-0.015)(t) = 2000sec

I'm not really sure of the answer as it seems to be quite impractical :)P).
Thanks.
~Troubadour

You missed the obvious solution. If you apply an accelerating force of 2000N for 15secs, how long do you have to apply a decelerating force of 15N to return to rest?

 

[HalfLostAndSearching]

I can confirm that your attempt at solving the problem is correct. However, the answer you have obtained may seem impractical because it does not factor in other variables such as air resistance, rolling resistance, and other external forces that may affect the motion of the car. In reality, the car would continue to slow down even after the engine is turned off and may not come to a complete stop after exactly 2000 seconds. Additionally, the assumption of a constant frictional force of 15N may not be accurate as friction can vary depending on the surface and other factors. Therefore, it is important to consider all factors and variables when making calculations and predictions in science.