# Mechanics – A car

• May 3rd 2006, 09:19 AM
Natasha1
Mechanics – A car
Any help with this would be most appreciated. Thanks in advance

The EMF (Effective Motive Force) for a car is the engine driving force minus friction and air resistance.

a)A car of mass 1000Kg travels from rest with constant acceleration to 25 m/s in 30 seconds. Use the impulse-momentum relationship to find the EMF?

b)Now suppose that instead the EMF is made by the driver to increase uniformly during the 30 seconds (by gradually increasing foot pressure on the accelerator). Use the impulse-momentum relationship to find the final speed reached?

c)Compare a) and b)?
• May 3rd 2006, 12:05 PM
topsquark
Quote:

Originally Posted by Natasha1
Any help with this would be most appreciated. Thanks in advance

The EMF (Effective Motive Force) for a car is the engine driving force minus friction and air resistance.

a)A car of mass 1000Kg travels from rest with constant acceleration to 25 m/s in 30 seconds. Use the impulse-momentum relationship to find the EMF?

b)Now suppose that instead the EMF is made by the driver to increase uniformly during the 30 seconds (by gradually increasing foot pressure on the accelerator). Use the impulse-momentum relationship to find the final speed reached?

c)Compare a) and b)?

The impulse-momentum theorem is:
$\overline{\sum F} \Delta t = \Delta p$ where $\overline{\sum F}$ is the average net force applied to the object.

a) The acceleration is constant, so the net force applied is constant. I will call the net force F for simplicity.
$F \Delta t = \Delta p = m \Delta v$
$F = m \frac{ \Delta v}{ \Delta t} = 1000 \frac{25}{30} \, N$
Thus F = 833 N.

b) The acceleration is now a linear function of time over the first 30 seconds of motion. Thus the net force is also a linear function of time. Using the Calculus version of the impulse-momentum theorem:
$\int_{t_0}^t F \,dt = \int_{v_0}^v m \, dv$

The problem I am having here is that the acceleration is a linear function of time, and we don't have the constant: $a(t) = ct$, giving $F(t)=ma(t)=mct$.

In terms of c:
$\int_0^{30}mct \, dt = \int_0^v m \, dv$
$\frac{1}{2}mct^2|_0^{30} = mv|_0^v$
$450mc=mv$

So $v = 450c \, \, m/s$.

-Dan