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 metersb) 1.00 × 103 metersc) 5.05 × 102 metersd) 2.00 × 103 meterse) 1.00 × 102 meters

Answer 1

Vi = 0m/s
A = 99m/$s^{2}$
Vf = 445m/s 
T = 4.5s 
D = ? 

D = Vi(t) + 1/2(a)(t²)
D = 0m/s ( 4.5s) + 1/2(99m/s²)(4.5s)²
D = 1002 m
D = 1 x 10³ m