A drag racer crosses the finish line of a 400.0-m track with a final speed of 104 m/s. (a) Assuming constant acceleration during the race, find the racer’s time and the minimum co efficient of static friction between the tires and the road. (b) If, because of bad tires or wet pavement the acceleration were 30.0% smaller, how long would it take to finish the race?
Vf^2=2ad
solve for a.
But force=ma
or mu*mg=ma
or mu=a/g is minimum coefficient
if a is .7 of the original, then
distance=1/2 a t^2 or t= sqrt 2distance/a
so what is the square root of (1/.7) ?
To find the time and the minimum coefficient of static friction in part (a) and the time in part (b), we can use the equations of motion and the concept of uniform acceleration.
(a) First, let's find the time it took for the racer to cross the finish line using the equation of motion:
vf = vi + at
In this case, the initial velocity (vi) is 0 m/s since the racer started from rest. The final velocity (vf) is 104 m/s, and the distance (s) is 400.0 m. We need to find the acceleration (a) and the time (t).
Using the equation:
vf^2 = vi^2 + 2as
we can solve for acceleration:
a = (vf^2 - vi^2) / (2s)
substituting the values:
a = (104^2 - 0^2) / (2 * 400.0)
Now, we can find the time using the equation:
t = (vf - vi) / a
Substituting the values:
t = (104 - 0) / a
Now, let's calculate the time.
(b) In this case, the acceleration is 30.0% smaller, so the new acceleration (a') will be:
a' = a - (0.3 * a)
Let's calculate the time using the equation:
t' = (vf - vi) / a'
Substituting the values:
t' = (104 - 0) / a'
Now, let's calculate the new time.
To find the minimum coefficient of static friction between the tires and the road, we can use the following equation:
μs = (ma - mg) / (m * g)
where μs is the coefficient of static friction, m is the mass of the racer, a is the acceleration, and g is the acceleration due to gravity.
However, the mass is not given in the question, so we cannot calculate the minimum coefficient of static friction without that information.