A motorcycle starts from rest with constant acceleration of 1.5m/s^2 until it achieves velocity of 15m/s. The motorcycle decelerate until it stop 12s. Determine:
1) the deceleration of motorcycle
2) distance traveled by the motorcycle
question one does not make sense to me.
2) figure time it takes for the first portion
vf=vi+at solve for time, vf=15, vi=0
distanceFirst=avgvelocity*time=7.5*time
distance Second=avgvelocity*12
and avg velocity for each phase again is 15/2 m/s
To determine the deceleration of the motorcycle, we can use the formula:
acceleration = (final velocity - initial velocity) / time
Given that the motorcycle starts from rest and achieves a velocity of 15 m/s, and the total time is 12 seconds, we can calculate the deceleration:
acceleration = (15 m/s - 0 m/s) / 12 s
acceleration = 15 m/s / 12 s
acceleration = 1.25 m/s^2
So, the deceleration of the motorcycle is 1.25 m/s^2.
To determine the distance traveled by the motorcycle, we can use the formula:
distance = (initial velocity * time) + (0.5 * acceleration * time^2)
Since the motorcycle starts from rest, the initial velocity is 0 m/s. The time is 12 seconds, and we already know the deceleration is 1.25 m/s^2. Plugging these values into the distance formula, we can calculate the distance traveled:
distance = (0 m/s * 12 s) + (0.5 * 1.25 m/s^2 * (12 s)^2)
distance = 0 m + 0.5 * 1.25 m/s^2 * 144 s^2
distance = 0 + 0.5 * 1.25 m/s^2 * 144 s^2
distance = 0 + 0.5 * 1.25 m/s^2 * 144 s^2
distance = 0 + 0.5 * 1.25 m/s^2 * 144 s^2
distance = 0 + 0.5 * 1.25 m/s^2 * 144 s^2
distance = 0 + 0.5 * 1.25 m/s^2 * 144 s^2
distance = 0 + 0.5 * 1.25 m/s^2 * 144 s^2
distance = 900 meters
So, the distance traveled by the motorcycle is 900 meters.