# physics, acceleration, physics problem

• Sep 28th 2007, 01:40 PM
rcmango
physics, acceleration, physics problem
For a standard production car, the highest road-tested acceleration ever reported occured in 1993, when a Ford RS200 Evolution went from zero to 26.8 m/s (60 mi/h) in 3.275 s. Find the magnitude of the car's acceleration.
displayed by
m/s2
Repeat, for another car which goes from zero to 60 mi/h in 4.940 s.
calculated in m/s2

trying to understand all the variable involved. thanks for any help you can give here. appreciate it.
• Sep 28th 2007, 02:24 PM
TKHunny
These are all very similar. Get those three equations in your head. You will see them all.
• Sep 28th 2007, 03:09 PM
ticbol
Quote:

Originally Posted by rcmango
For a standard production car, the highest road-tested acceleration ever reported occured in 1993, when a Ford RS200 Evolution went from zero to 26.8 m/s (60 mi/h) in 3.275 s. Find the magnitude of the car's acceleration.
displayed by
m/s2
Repeat, for another car which goes from zero to 60 mi/h in 4.940 s.
calculated in m/s2

trying to understand all the variable involved. thanks for any help you can give here. appreciate it.

[SUVAT! :). Motion caused by..., or where acceleration is constant.]

Evolution went from zero to 26.8 m/s (60 mi/h) in 3.275 s. Find the magnitude of the car's acceleration, displayed by m/s^2

V final, Vf = Vo +at -----------(i)

Given:
Vo = 0
Vf = 26.8 m/sec
t = 3.275 sec

Find a.

26.8 = 0 +a(3.275)
a = 26.8 /3.275 = 8.183 m/sec/sec -------------answer.

Repeat, for another car which goes from zero to 60 mi/h in 4.940 s.
calculated in m/s2

Vo = 0
Vf = 60mi/hr = 26.8m/sec
t = 4.94 sec

Vf = Vo +at
26.8 = 0 +a(4.94)
a = 26.8 /4.94 = 5.425 m/sec/sec --------------answer.
• Sep 29th 2007, 03:50 AM
topsquark
Quote:

Originally Posted by TKHunny
These are all very similar. Get those three equations in your head. You will see them all.

Actually, I find that the following set of four equations is slightly more useful:
$\displaystyle x = x_0 + v_0t + \frac{1}{2}at^2$

$\displaystyle v = v_0 + at$

$\displaystyle v^2 = v_0^2 + 2a(x - x_0)$

This last equation is not strictly necessary, but including it means that students do not need to solve multiple equations in multiple unknowns (in 1-D motion, anyway.)
$\displaystyle x = x_0 + \left ( \frac{v_0 + v}{2} \right ) t$

-Dan