calculating speed of an accelerating vehicle

Question

When a vehicle accelerates from a dead stop and continues to accelerate, calculate its approximate speed after traveling 860 feet based upon the following published reports of the vehicle's performance: (a) it can go from zero to 60 mph in 7.4 seconds; and (b) its speed at a quarter mile is 91 mph. Also, how many seconds does it take the vehicle to travel 860 feet?

Answer 1

Let's show that there are 2 different accelerations in part a and b. Here, it helps to keep your units of measure consistent, so lets stick with mph. Any distance is in miles and any time in hours.

From part a) we can work out the car's time average acceleration. This will be in miles per hour per hour.

a = (v - 0)/t = (60 - 0)[mph] / 0.002056 [hours] = 29,200 mph/hr

The distance it must have traveled to do this is given by the following kinematic equation:

v^2 = 0^2 + 2a*d means that d = 0.06167 miles = 325.6 ft

As mentioned already, for part b) infact, the car must not have the same acceleration as in part a. You can show this is true by plugging in the values to the kinematics equation here, and finding that a (acceleration) is different from what we have above:

v^2 = 0^2 + 2a*d, for d = 0.25 miles and v = 91 mph implies a = 16560 mph/hr

Hmm.... so what does this mean. Perhaps we can assume that section b happens at a later stretch of the race, and has a different acceleration. Then it would be:

v_f^2 = v_i^2 + 2a*d, for d = 0.25 miles and v_f = 91 mph and v_i = 60 implies a = 9362 mph/hr

Notice here there would be 2 different accelerations, as there are 2 different sections of time. You should ask the question writer whether this is what they meant.

If so, then use this equation to solve for time as a function of distance for part b. Then add this time to the time in part a.

d = V_i*t + 1/2*a*t^2