Question

6. One minute after takeoff, a rocket carrying the space shuttle into outer space reaches a speed of 447 m/s.

Answer 1

Answer:

acceleration of the rocket is given as

$a = 7.45 m/s^2$

Explanation:

As we know that rocket starts from rest and then reach to final speed of 447 m/s after t = 1 min

so we have

$v_i = 0$

$v_f = 447 m/s$

$t = 1 min = 60 s$

so we have

$a = \frac{v_f - v_i}{t}$

$a = \frac{447 - 0}{60}$

$a = 7.45 m/s^2$