Simple motion problem

1. Dec 8, 2006

ffww192

A car can accelerate uniformaly to 120 mi/h in 30 s. It is max braking rate cannot exceed 0.7g. What is the minimum time required to go 1/2 mile? Assume it begins and ends at rest. (Hint: a graph of velocity vs. tme can be helpful).

I am a high school teacher and one of my students gave me this problem to solve. I have tried my best to solve it but no luck of getting it. Any help is appriciated.

2. Dec 8, 2006

Ja4Coltrane

easy: acceleration is 120 over 30 seconds=4mi/((hr)(seconds))
but you need to convert it into a better unit like mi/sec^2
do this by dividing by 3600
then you have .00111mi/sec^2
(.03333mi/sec)^2=2(.0011)d
so
.5 mi=.5at^2
t=300 sec

3. Dec 8, 2006

sorry 30 sec

4. Dec 9, 2006

OlderDan

The car can accelerate uniformly for some distance and brake uniformly for some distance. The total distance must be d = 1/2 mile. Ja4Coltrane showed that the acceleration can be computed as 4mi/h/s, which is best converted to units of ft/s² with d expressed in feet. The maximum velocity of the car can be expressed in terms of the acceleration and the distance traveled during acceleration, s, as (starting from rest)

v² = 2as

The car then declerates from maximum velocity at 0.7g to a final velocity of zero in a distance of (d - s)

0 = v² - 2*0.7g(d - s)
v² = 2*0.7g(d - s)

Combining these equations gives

2as = 2*0.7g(d - s)

which can be solved for s. When s is found, it can be used to find the maximum velocity from either equation for v². The average velocity during both acceleration and deceleration is half the maximum velocity. Assuming the maximum velocity is less that 120mph, half the maximum velocity is the average velocity for the whole trip. The time is the half mile distance divided by the average velocity.

If it should happen that the computed maximum velocity is greater than 120mph (it's not) and you assume the car can go no faster than 120mph, you would have to break the trip up into three parts and compute the three times separately.