For the first 4 seconds of a race, a sprinter accerlates at a rate of 3m/sec^2.

Question

He then continues to run at the constant speed that he has attained for the rest of the race.

A) Write a piecwise function to express the sprinter's velocity v(t) as a function of time. 

B) Find v(2), v(4), and v(6).

C) Write a piecewice function to express the sprinter's position s(t) as a function of time. 

D) How far does the sprinter run during the first 4 seconds of the race?

E) How long will it take the sprinter to run 100m?

Answer 1

A) v(t) = 3t, 0 <= t <= 4

              12, t > 4

where v is in m/s

 

B) v(2) = 3(2) = 6 m/s

    v(4) = 3(4) = 12 m/s

    v(6) = 12 m/s

 

C) x = integral[v(t)]dt

       = (3/2)t^2 + x0 0 <= t <= 4

       = 12 + x1, t > 4

 

where x0 and x1 are constants. Assuming that the initial position of the race is x = 0, x0 = 0. Then, x(t) - (3/2)t^2. So, x(4)  = (3/2)(4^2) = (3/2)*16 = 24. So, v(4) = 12(4) + x1 = 24, matching with the second piece of the position function. v(4) = 48 + x1 = 24. So, x1 = -24.

 

x(t) = (3/2)t^2, 0 <= t <= 4

          12t - 24, t > 4

 

D) During the first 4 seconds of the race, the sprinter runs x(4) = (3/2)(4^2) = 24 m.

4) He takes 4 seconds to run 24 m. He needs to run the other 100 m - 24 m = 76 m at constant speed.

76 m = (12 m/s)t

t = 6.33 sec.

So, the total time to run 100 m is 4s + 6.33s = 10.33s

 

Answer 2

This is a fairly simple kinematic equation problem, which has been made difficult with all the sub-problems.

 

We can find the instantaneous velocity of an object that is accelerating using the formula

 

v = v_i + at

 

This is actually the velocity as a function of time. So let's plug in our values and we find the first part of the piecewise function (the initial velocity is 0 because this is a race, and everybody starts from rest).

 

v(t) = at = 3t

 

For times after 4 seconds, the sprinter has a constant velocity. We find this velocity by plugging in t=4 into the previous equation.

 

v(t>4) = 3*4 = 12

 

So our piecewise function is 

 

3t   for t<4

12  for t>=4

 

Now for part B, we should use our piecewise function to find the values at the given times

 

v(t=2) = 3*2 = 6 m/s

v(4) = 12 m/s

v(6) = 12 m/s

 

PART C

This is slightly more complicated. We find the position as a function of time using

 

s(t) = v₀t + (1/2)at^t

 

This works for both the times when the runner is accelerating and when he is running at a constant speed. For the constant speed part, we need to make an adjustment.

 

For 0<t<4 we have

 

s(t) = 0 + (1/2)3t²

 

Once the runner has completed this step, we need to know both the position and velocity at 4 seconds. Using our previous position and velocity functions, we find that his position is 24 m, and his velocity is 12 m/s.

 

His position function after 4 s needs to consider is position at 4s, then go from there. So it should have a value of 24 when t = 4. We shift the velocity like

 

s(t>4) = 24 + v(t - 4)

 

This should make sense because when t=4, the position is 24 m, and this is an equation of a line. So our piecewise function is

 

s(0<4) =  1.5t²

s(t>4) = 24 + 12(t - 4)

 

For part D, we have actually already found that distance. It is the 24 m that we found earlier. 

 

For part E, we need to be careful again. We know that it took 4 seconds for the runner to go 24 m. Now how long does it take to go the extra 76 m? The runner has a constant velocity of 12 m/s after 4 seconds, so what we want to know is how long it take him to run 76 m at a rate of 12 m/s. 

 

76 m / (12 m/s) = 6.333 seconds

 

So the total time is 4 + 6.333 = 10.3 s

 

I hope this helps! Please ask any questions if you have them.

Answer 3

a X t =velocity

 

(0,4) v(t) = 3t

 

(4, rest of race) v(t) = 12m/sec

 

B) v(2) = 6m/sec  v(4) = 12m/sec  v(6) = 12m/sec

 

(1/2)at² = s(t)

 

C) (0,4) s(t)=(1/2)at²

    (4, rest of race) s(t) = 12t

 

D) (0,4) s =(1/2)X3X16 = 24meters

 

E) 100-24=76 meters at 12m/sec   t=4 +(76/12) seconds to run 100meters