Question

A rocket 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

Answer 1

Answer: 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 $99.0m/s^2$

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

$v^2-u^2=2as$

where, v = final velocity = 445m/s

u = initial velocity = 0m/s

a = acceleration = $99m/s^2$

s = distance = ? m

Putting values in above equation, we get:

$(445)^2-(0)^2=2(99)s\\\\s=1\times 10^3meters$