Answer:
[tex]62.7 m/s^2[/tex]
Explanation:
Assuming the dragster accelerates at constant rate, the equation that describes the distance travelled by the dragster after time t is
[tex]d(t) = ut + \frac{1}{2}at^2[/tex]
where
u is the initial velocity
a is the acceleration
If we assume that the dragster starts from rest,
u = 0
And we also know that at t = 3.58 s, the distance covered is
d = 402 m
Solving the formula for a, we find the acceleration:
[tex]d=\frac{1}{2}at^2\\a=\frac{2d}{t^2}=\frac{2(402)}{(3.58)^2}=62.7 m/s^2[/tex]
Answer:
[tex]a = 62.7 m/s^2[/tex]
Explanation:
As we know that the distance moved by him is
[tex]d = 402 m[/tex]
the total time for which he is accelerating is given
t = 3.58 s
now we know that
[tex]d = \frac{1}{2}at^2[/tex]
[tex]402 = \frac{1}{2}a(3.58^2)[/tex]
[tex]a = 62.7 m/s^2[/tex]
So he must run with the above acceleration so he will move the distance in shortest possible time
