Question

Can you help me with this question.

Answer 1

Given:

he initial velcicty of the runner is

`v_(ir)=\text{ 0 m/s}`

The final velocity of the runner is

`v_(fr)=\text{ 2.6 m/s}`

The time taken is t = 2.1 s

The initial velocity of the motorcycle is

`v_(im)=\text{ 37 m/s}`

The final velocity of the motorcycle is

`v_(fm)=\text{ 44 m/s}`

Required:

Acceleration of the motorcycle.

The difference in distance between runner and motorcycle.

Step-by-step explanation:

cceleration of the runner is

`\begin{gathered} a_r=(v_(fr)-v_(ir))/(t) \\ =(2.6-0)/(2.1) \\ =1.238\text{ m/s}^2 \end{gathered}`

Acceleration of the motorcycle is

`\begin{gathered} a_m=(v_(fm)-v_(im))/(t) \\ =(44-37)/(2.1) \\ =3.33\text{ m/s}^2 \end{gathered}`

The distance travelled by the runner is

`\begin{gathered} S_r=v_(ir)t+(1)/(2)a_rt^2 \\ =0*2.1+(1)/(2)*1.238*(2.1)^2 \\ =2.73\text{ m} \end{gathered}`

The distance travelled by motorcycle is

`\begin{gathered} S_m=v_(im)t+(1)/(2)a_mt^2 \\ =37*2.1+(1)/(2)*3.33*(2.1)^2 \\ =77.7+7.34 \\ =85.04\text{ m} \end{gathered}`

Thus, the motorcycle travels more distance than the runner.

The

difference between the distance travelled is

`\begin{gathered} \Delta S=S_m-S_r \\ =85.04-2.73 \\ =82.31\text{ m} \end{gathered}`

Final Answer:

The acceleration of the runner is 1.238 m/s^2

The acceleration of the motorcycle is 3.33 m/s^2

The difference in distance between runner and motorcycle is 82.31 m.