Question

On takeoff, a rocket accelerates from rest at a rate of 50.0 m/s^2 for exactly 1 minute. The rocket's speed after this minute of steady acceleration will be _____ m/s. (a) 50 (b) 500 (c) 3.00 x 10^3 (d) 3.60 x 10^3 (e) none of these

Answer 1

Answer: Option c.

Step-by-step explanation:

You need to use the following Kinematic equation:

`V_f=V_0+at`

Where
`V_f` is the final velocity,
`V_0` is the initial velocity,
`a` is the acceleration and
`t` is the time.

In this case, you can identify that:

`V_0=0\ (it\ accelerates\ from\ rest)\\\\a=50.0\ (m)/(s^2)\\\\t=1\ min=60\ s`

Therefore, substituting these values into the formula, you get:

`V_f=0+(50.0\ (m)/(s^2))(60\ s)\\\\V_f=3,000\ (m)/(s)\\\\V_f=3.00*10^3\ (m)/(s)`