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A powerful motorcycle can accelerate from rest to [tex]27 \, \text{m/s}[/tex] in only [tex]4.3 \, \text{s}[/tex]. How far does it travel in that time? (Express the numeral of your answer in SI units.)

Answer :

The motorcycle travels a distance of 58.05 meters in 4.3 seconds.

To calculate the distance traveled, we can use the kinematic equation:

\[d = \frac{1}{2}at^2\]

Where:

d = distance traveled

a = acceleration

t = time

Given that the motorcycle starts from rest, its initial velocity (u) is 0 m/s. The final velocity (v) is given as 27 m/s, and the time taken (t) is 4.3 seconds.

To find the acceleration (a), we can use the formula:

\[a = \frac{v-u}{t}\]

Substituting the values, we get:

\[a = \frac{27 - 0}{4.3} = 6.279 \, \text{m/s}^2\]

Now, we can plug the values of acceleration and time into the distance formula:

\[d = \frac{1}{2} \times 6.279 \times (4.3)^2 = 58.05 \, \text{meters}\]

Therefore, the motorcycle travels a distance of 58.05 meters in 4.3 seconds.

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