Arocket initially at rest accelerates at a rate of 99.0 meters/second2. calculate the distance covered by the rocket if it attains a final velocity of 445 meters/second after 4.50 seconds.
a) 2.50 × 102 meters
b) 1.00 × 103 meters
c) 5.05 × 102 meters
d) 2.00 × 103 meters
e) 1.00 × 102 meters

The correct answer is option b.

Explanation: We are given that the rocket is at rest initially final velocity is 445m/s.

The acceleration of the rocket is

To calculate the distance of rocket, we use third equation of motion, which is:

where, v = final velocity = 445m/s

u = initial velocity = 0m/s

a = acceleration =

s = distance = ? m

Putting values in above equation, we get:

the answere is B

Explanation:

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.

i really dont know but i will try okay.

so i think u would just add all them up and get your answer..