Maggie:
1/2 a*2.2^2 + 2.2a*8.4 = 100
a = 4.78
Judy:
1/2 a*2.9^2 + 2.9a*7.7 = 100
a = 3.77
In a 100 m race, Maggie and Judy cross the finish line in a dead heat, both taking 10.6 s. Accelerating uniformly, Maggie took 2.20 s and Judy 2.90 s to attain maximum speed, which they maintained for the rest of the race. What was Maggie's acceleration?
4 answers
Thanks so much!!! If you have time can you help me figure out their max speeds?
Fot max speed, multiply the acceleration rate by the length of time that person accelerated
I would like to share the method I was able to use to find the answer that Steve posted above.
The answer is correct (I plugged in numbers and the answer in the back of the book matches what I got), but I wanted to find how exactly those equations were found. What I gathered is the two equations used were,
(i) V = u + at
and
(ii) x = ut + 1/2 at^2, where u is initial velocity.
Since the total distance is known, and the fact that (ut) and (1/2at^2) each are distances, we just have to find (ut) in terms of a. We want our equation to add up to 100, so we have to find how initial velocity can be used since they start from rest. Upon closer inspection, there are two initial velocities. One from the start and the next when the top acceleration is reached. Using (i), u goes to zero,
so Vmax = at. We may plug this into equation (ii) because the max velocity is the initial velocity for the start of constant velocity.
x = (Vmax)(T) + 1/2at^2, where (T) is the time of constant velocity and (t) is the time of acceleration.
Substituting in at for Vmax, we arrive at,
x = (at)(T) + 1/2at^2.
The answer is correct (I plugged in numbers and the answer in the back of the book matches what I got), but I wanted to find how exactly those equations were found. What I gathered is the two equations used were,
(i) V = u + at
and
(ii) x = ut + 1/2 at^2, where u is initial velocity.
Since the total distance is known, and the fact that (ut) and (1/2at^2) each are distances, we just have to find (ut) in terms of a. We want our equation to add up to 100, so we have to find how initial velocity can be used since they start from rest. Upon closer inspection, there are two initial velocities. One from the start and the next when the top acceleration is reached. Using (i), u goes to zero,
so Vmax = at. We may plug this into equation (ii) because the max velocity is the initial velocity for the start of constant velocity.
x = (Vmax)(T) + 1/2at^2, where (T) is the time of constant velocity and (t) is the time of acceleration.
Substituting in at for Vmax, we arrive at,
x = (at)(T) + 1/2at^2.
1 answer