Acceleration question?

Question

If a car takes 7 seconds to go from 0-60, how far would it had traveled by the time it reaches 60 mph? Probably something to do with integrals (or was that derivative) and the distance formula. How would one go solving this problem? Thanks.

Answer 1

x=x (initial)+ (integral)V*dt or v^2 - v (initial)^2 = 2a(x-x (initial))

x initial is 0 (your starting point)
dt=7 sec
acceleration is dv/dt, or 60 mph/7 sec, but this needs converted
into ft per sec

so 60*5280 = 316,000 and there are 3600 sec in 1 hr

316000/3600 = 88
a = 88/7 ~ 12.57

so 88^2 = 7744
(7744 - 0)/ 2(12.57) = x - x initial and x initial is 0 so set = just x

7744/25.14 = 308.04 ft

Answer 2

The formulas which connect the distance S, the final speed V, a constant acceleration A and the time T are V = A x T, and V^2 = 2 x A x S, as long as everything is in the same units. So the first thing is to combine the two equations to eliminate A and get S = V x T / 2, and the next is to say 88 ft/sec instead of 60 mph. Then S = 88 * 7 / 2 = 308 feet.

But of course a real car's acceleration is higher at the start than when it has speeded up, so the actual distance will be further than that.

Answer 3

This is not a static physic problem anymore, this became Dynamic Movement becasue the acceleration is not cero.
IF you had Space= Velocity * Time this wont work you need to go one step back in the developing of this formula where you can find the integrals. I do not have the formula in my mind anymore since it has been a while since i left college.
I hope it helps....