Determine the distance traveled by a particle whose initial velocity is 48km/h. The particle accelerate uniformly at Tate of 1.8m/s and attained a velocity of 72km/h?
Vo = 48km/h = 48,000m/3600s = ___ m/s.
V = 72km/h =
V^2 = Vo^2 + 2a*d. Solve for d.
To determine the distance traveled by a particle, we can use the equation of motion for uniformly accelerated motion.
The equation is:
s = (v^2 - u^2) / (2a)
Where:
- s represents the distance traveled
- v represents the final velocity
- u represents the initial velocity
- a represents the acceleration
In this case, the initial velocity (u) is 48 km/h, the final velocity (v) is 72 km/h, and the acceleration (a) is 1.8 m/s^2.
First, we need to convert the initial and final velocities to meters per second (m/s):
48 km/h = (48 * 1000) / (60 * 60) = 13.333 m/s
72 km/h = (72 * 1000) / (60 * 60) = 20 m/s
Now, we can substitute the values into the equation:
s = (20^2 - 13.333^2) / (2 * 1.8)
s = (400 - 177.78) / 3.6
s = 222.22 / 3.6
s = 61.728 m
Therefore, the distance traveled by the particle is approximately 61.728 meters.