Rocket Acceleration?

Question

A rocket accelerates uniformly from rest to a speed of 4.2 x 10^3 mph in 125 sec.. Over what distance does the rocket accelerate? (Please explain how you got your answer)

Answer 1

hi,

u=o m/sec

v=4.2x1000/3600m/sec

=1.167m/sec

t=125 sec

therefore a=v-u/t=1.167-0/125=0.0093m/sec^2

v^2-u^2=2as

1.167^2-0=0.0186s

therefore s=73.21m(approx)

Answer 2

According to your question, in one hour, the rocket covers a distance of 4.2 x 10^3 mph. Since, 1 mile=1.6 km(approx), in one hour it should cover 4.2 x 1.6 x 10^3 km = 42 x 16 x 10^5 km. 1km/h=1000m/60x60 secs = 5/18 m/s. therefore, in the speed of the rocket is 42 x 16 x 10^3 x 5/18 m/s.(converting km/h to m/s). you should get 7 x 10^6m/s as the speed. Use the formula for acceleration a=v-u/t. v=final velocity u=initial velocity. a=7 x 10^6-0/125=5.6 x 10^4m/s^2, which is the acceleration. Since, the rocket accelerates from rest, the entire distance 4.2 x 10^3 mph is the distance over which the rocket accelerates.

Answer 3

Since it accelerates uniformly...

1.) Convert 125 seconds into hours.

2.) Divide the final speed by the answer above.

Done.