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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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