Question

An airplane covers a straight-line distance of 8.13 km in 33.5 s, during which time it has a constant forward acceleration of 4.6 m/s2.1. what is the speed at the first begining of 33.5 s.2.what is the speed at the end of the 33.5 s.

Answer 1

Given data

*The given distance is s = 8.13 km = 8130 m

*The given time is t = 33.5 s

*The given acceleration is a = 4.6 m/s^2

(1)

The formula for the speed at the first beginning of 33.5 seconds is given as

`s=ut+(1)/(2)at^2`

Substitute the known values in the above expression as

`\begin{gathered} 8130=u(33.5)+(1)/(2)(4.6)(33.5)^2 \\ u=165.6\text{ m/s} \end{gathered}`

Hence, the speed at the first beginning of 33.5 s is u = 165.6 m/s

(2)

The formula for the speed at the end of the 33.5 s is given as

`v=u+at`

Substitute the known values in the above expression as

`\begin{gathered} v=(165.6)+(4.6)(33.5)_{} \\ =319.7\text{ m/s} \end{gathered}`

Hence, the speed at the end of the 33.5 s is v = 319.7 m/s