Question

a motorcycle travels west increasing its speed from 6.5m/s to 9.2m/s through a distance of 48m. what was the motorcycle's acceleration?

Answer 1

The rate of acceleration of the motorcycle is 0.442 $m/s^2$.

Given the following data:

• Initial speed = 6.5 m/s  

• Initial speed = 9.2 m/s  

• Distance = 48 meters.

To find the rate of acceleration of the motorcycle, we would use the third equation of motion;

$V^2 = U^ + 2aS$

Where:

• V is the final speed.

• U is the initial speed.

• a is the acceleration.

• S is the distance covered.

Substituting the given parameters into the formula, we have;

$9.2^2 = 6.5^2 + 2a(48)\\\\84.64 = 42.25 + 96a\\\\96a = 84.64 - 42.25\\\\96a = 42.39\\\\a = \frac{42.39}{96}$

Acceleration, a = 0.442 $m/s^2$

Therefore, the rate of acceleration of the motorcycle is 0.442 $m/s^2$.

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