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

The rocket will cover $1 \times 10^3\rm \ m$ distance in 4. 5 s. Acceleration can be defined as the change in velocity.

What is acceleration?

Acceleration can be defined as the change in speed or the direction of the object.

From kinamatic equation:

$D = v_{t} +\dfrac 12at^2$

Where,

$D$ - final velocity =  445 m/s

$v_0$ -  initial valocity = 0 m/s

$a$ - acceleration = 99. 0 m/s²

$t$ - time =  4. 50 s

Put the values in the formula,

$D = 0\times ( 4.5) + \dfrac12\times (99)(4.5)^2\\\\D = 1002 {\rm \ m}\\\\D = 1 \times 10^3\rm \ m$

Therefore, the rocket will cover $1 \times 10^3\rm \ m$ distance in 4. 5 s.

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