Question

A powerful motorcycle can accelerate from rest to 28 m/s in 4 seconds. What is its average acceleration? How far does it travel in that time?​

Answer 1

Answer:-

• Initial velocity (u) = 0
• Final velocity (v) = 28m/s
• Time taken (t) = 4sec
• Acceleration (a) = ?
• Distance travelled (s) = ?

$\Large{\bf{\blue{\underline{\underline{\pink{Solution}}}}}}$

According to the first equation of motion

$\begin{gathered}\begin{gathered}\implies\sf v=u+at \\ \\ \\ \implies\sf 28=0+a\times{4} \\ \\ \\ \implies\sf 28=4a \\ \\ \\ \implies\sf a=\cancel\dfrac{28}{4}=7m/s^2\end{gathered}\end{gathered}$

According to the third equation of motion

$\begin{gathered}\begin{gathered}\implies\sf v^2-u^2=2as \\ \\ \\ \implies\sf (28)^2-(0)^2=2\times{7}\times{s} \\ \\ \\ \implies\sf 784=14s \\ \\ \\ \implies\sf s=\cancel\dfrac{784}{14}=56m\end{gathered}\end{gathered}$

Hence, acceleration of motorcycle is 7m/s² and distance travelled by it is 56m

Answer 2

Answer:-

Initial velocity (u) = 0

Final velocity (v) = 28m/s

Time taken (t) = 4sec

Acceleration (a) = ?

Distance travelled (s) = ?

\Large{\bf{\blue{\underline{\underline{\pink{Solution}}}}}}

Solution

According to the first equation of motion

\begin{gathered}\begin{gathered}\begin{gathered}\implies\sf v=u+at \\ \\ \\ \implies\sf 28=0+a\times{4} \\ \\ \\ \implies\sf 28=4a \\ \\ \\ \implies\sf a=\cancel\dfrac{28}{4}=7m/s^2\end{gathered}\end{gathered}\end{gathered}

⟹v=u+at

⟹28=0+a×4

⟹28=4a

⟹a=

4

28

=7m/s

2

According to the third equation of motion

\begin{gathered}\begin{gathered}\begin{gathered}\implies\sf v^2-u^2=2as \\ \\ \\ \implies\sf (28)^2-(0)^2=2\times{7}\times{s} \\ \\ \\ \implies\sf 784=14s \\ \\ \\ \implies\sf s=\cancel\dfrac{784}{14}=56m\end{gathered}\end{gathered}\end{gathered}

⟹v

2

−u

2

=2as

⟹(28)

2

−(0)

2

=2×7×s

⟹784=14s

⟹s=

14

784

=56m

Hence, acceleration of motorcycle is 7m/s² and distance travelled by it is 56m