Question

A rocket initially traveling straight up at a speed of acceleration ? 20.0 m/s undergoes constant acceleration in the direction of travel to a speed of 30.0 m/s over a distance of 40.0 m What is the magnitude of the rocket's

Answer 1

Given data

*The given initial speed of the rocket is u = 20.0 m/s

*The given final speed of the rocket is v = 30.0 m/s

*The given distance is s = 40.0 m

The formula for the magnitude of the rocket's acceleration is given by the kinematic equation of motion as

`\begin{gathered} v^2=u^2+2as \\ a=(v^2-u^2)/(2s) \end{gathered}`

Substitute the known values in the above expression as

`\begin{gathered} a=((30.0)^2-(20.0)^2)/(2(40)) \\ =6.25m/s^2 \end{gathered}`

Hence, the magnitude of the acceleration of the rocket is a = 6.25 m/s^2