A motor cyclist traveling at 30 m/s accelerates uniformly for 5 seconds to v= 40 m/s. H then maintains a constant velocity for the next 4 minutes. How far did the motor cyclist travel throughout the entire motion?
a = (Vf - Vo) / t,
= (40 - 30) / 5 = 2 m/s^2.
4 min * 60 s/min = 240 s.
d = Vo*t + 0.5*at^2,
= 30*240 + 0.5 * 2 * (240)^2,
= 7200 + 57600 = 64800 m = 64.8km.
A more practical approach
4 min. = 240 s.
d=V*t=40 m/s * 240 s = 9600 m = 9.6km.
To find the total distance traveled by the motorcyclist, we need to consider the distance covered during the acceleration phase and the distance covered during the constant velocity phase.
1. Distance covered during acceleration:
The motorcyclist initially travels at a velocity of 30 m/s and accelerates uniformly for 5 seconds to reach a final velocity of 40 m/s. We can use the formula for average acceleration:
a = (v - u) / t,
where,
a = average acceleration,
v = final velocity,
u = initial velocity,
t = time taken.
Substituting the given values:
a = (40 - 30) / 5 = 2 m/s^2.
Now, we can use the equation of motion to find the distance covered during the acceleration phase:
s = ut + (1/2)at^2,
where,
s = distance covered,
u = initial velocity,
t = time taken,
a = acceleration.
Substituting the values:
s = (30 * 5) + (1/2) * 2 * (5^2) = 150 + 2 * 25 = 150 + 50 = 200 meters.
2. Distance covered during constant velocity:
The motorcyclist maintains a constant velocity for 4 minutes, which is equal to 4 * 60 = 240 seconds. The distance covered during this phase can be calculated using the formula:
s = vt,
where,
s = distance covered,
v = velocity,
t = time taken.
Substituting the values:
s = 40 * 240 = 9600 meters.
3. Total distance traveled:
The total distance traveled by the motorcyclist is the sum of the distances covered during the acceleration phase and the constant velocity phase:
Total distance = 200 + 9600 = 9800 meters.
Therefore, the motorcyclist traveled a total distance of 9800 meters throughout the entire motion.