Question

If a body which is initially at rest speeds up to 8m/s in 5s then to 16m/s in 10s and to 24 m/s in 15s what is its average speed? Consider the acceleration to be constant

Answer 1

Hey, there!

Here's your answer.

If you draw the velocity time graph of the given data then it will be a straight line (as acceleration is constant) passing through origin (as initial speed is zero).

(See the attachment)

The total distance travelled by the body = area under the v-t curve = area of the triangle (here)

$= \frac{1}{2} \times 24 \times 15 \: m \\ \\ = 180 \: m$

$Average \: speed = \frac{total \: distance}{total \: time} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \: \frac{180}{15} = 12m {s}^{ - 1}$

Hope that helps.

Thanks.

Answer 2

Answer:

Here's your answer.

If you draw the velocity time graph of the given data then it will be a straight line (as acceleration is constant) passing through origin (as initial speed is zero).

The total distance travelled by the body = area under the v-t curve = area of the triangle (here)

\begin{gathered} = \frac{1}{2} \times 24 \times 15 \: m \\ \\ = 180 \: m\end{gathered}

=

2

1

×24×15m

=180m

\begin{gathered}Average \: speed = \frac{total \: distance}{total \: time} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \: \frac{180}{15} = 12m {s}^{ - 1} \end{gathered}

Averagespeed=

totaltime

totaldistance

=

15

180

=12ms

−1

Hope that helps.

Thanks.