At a football tryout, a player runs a 40âyard dash in 4.25 seconds. If he reaches his maximum speed at the 16âyard mark with a constant acceleration and then maintains that speed for the remainder of the run, determine his acceleration over the first 16 yards, his maximum speed, and the time duration of the acceleration.
Answers
Answered by
12
Hey buddy,
◆ Answer-
t = 2.43 s
v = 13.16 m/s
a = 5.41 m/s^2
◆ Explanation-
Let t be time duration of acceleration, v be maximum velocity and a be constant acceleration.
During acceleration -
(v+u) / 2 = s/t
v + 0 = 2×16/t
v = 32/t
For remaining motion-
s = vt
40 - 16 = v (4.25-t)
24 = v(4.25-t)
24×t/32 = 4.25 - t
7t/4 = 4.25
t = 2.43 s
Maximum velocity is-
v = 32/t
v = 32/2.43
v = 13.16 m/s
Constant acceleration is -
a = (v-u) / t
a = 13.16 / 2.43
a = 5.41 m/s^2
Hope this helps...
◆ Answer-
t = 2.43 s
v = 13.16 m/s
a = 5.41 m/s^2
◆ Explanation-
Let t be time duration of acceleration, v be maximum velocity and a be constant acceleration.
During acceleration -
(v+u) / 2 = s/t
v + 0 = 2×16/t
v = 32/t
For remaining motion-
s = vt
40 - 16 = v (4.25-t)
24 = v(4.25-t)
24×t/32 = 4.25 - t
7t/4 = 4.25
t = 2.43 s
Maximum velocity is-
v = 32/t
v = 32/2.43
v = 13.16 m/s
Constant acceleration is -
a = (v-u) / t
a = 13.16 / 2.43
a = 5.41 m/s^2
Hope this helps...
Similar questions