Final answer:
The distance covered by a motorcycle accelerating from 72 km/h to 88 km/h at a rate of 85m/s² is approximately 1.165 meters after converting the velocities from km/h to m/s.
Step-by-step explanation:
The question deals with calculating the distance covered by a motorcycle when it accelerates from one velocity to another. To solve this problem, we need to use the kinematic equations that relate acceleration, initial velocity, final velocity, and distance.
Firstly, we need to convert the speeds from kilometers per hour (km/h) to meters per second (m/s) as the acceleration is given in meters per second squared (m/s2). The conversion for velocity from km/h to m/s is done by multiplying by ⅓ m/s per km/h. So, the initial velocity (u) is 72 km/h, which equals 20 m/s, and the final velocity (v) is 88 km/h, which equals 24.444 m/s.
We can use the equation v2 = u2 + 2as, where 'a' is the acceleration and 's' is the distance. Rearranging this equation to solve for 's' gives us s = (v2 - u2)/2a.
Plugging the values into the equation:
s = ((24.444)2 - (20)2)/(2 × 85) = (598.13 - 400) / 170 = 198.13 / 170 ≈ 1.165 m
Therefore, the motorcycle traveled approximately 1.165 meters when it accelerated from 72 km/h to 88 km/h.