If the velocity of a particle decreases at 18 m/s^2 from when it is released at 100 m/s till a final velocity of 4 m/s is reached, what distance is traveled in 36 s.
The particle is released at 100 m/s and its velocity decreases at 18 m/s^2 till a velocity of 4 m/s is reached. The distance traveled in 36 s has to be determined.
First determine the distance traveled while the velocity of the particle is decreasing. The rate of decrease in velocity is 18 m/s^2. Use the relation v = u + at, where u is the initial velocity, v is the final velocity and a is the rate of change of velocity. Substituting the values given, 4 = 100 - 18*t
=> t = 96/18 = 5.33 s
The distance traveled in this duration is D = (100^2 - 4^2)/2*18 = 277.33 m
The time left for the particle to travel at the constant velocity of 4 m/s is 36 - 5.33 = 30.67 s
The distance traveled in 30.67 s is 4*30.67 = 122.67 m
The total distance traveled in 36 s is 122.67 + 277.33 = 400 m
The particle travels 400 m in 36 s.