a motorcycle moves at an initial velocity of 40km/h and accelerates at a constant rate of 8.3 m/s^2 for a distance of 120m how fast is the motorcycle then moving?
vf^2 - vi^2 = 2ax
Need to change 40 kph to m/s
To find the final velocity of the motorcycle, we can use the equation:
v^2 = u^2 + 2as
Where:
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance
First, we need to convert the initial velocity from km/h to m/s:
Initial velocity (u) = 40 km/h = 40 * (1000/3600) m/s = 11.11 m/s (rounded to two decimal places)
Now we can substitute the values into the equation:
v^2 = (11.11 m/s)^2 + 2 * 8.3 m/s^2 * 120 m
v^2 = 123.21 m^2/s^2 + 1992 m^2/s^2
v^2 = 2115.21 m^2/s^2
Now we can calculate the square root of both sides to find v:
v = √(2115.21 m^2/s^2)
v ≈ 46.02 m/s (rounded to two decimal places)
Therefore, the motorcycle is then moving at approximately 46.02 m/s.