A particle travelling in a straight line with constant acceleration and starting from rest covers 36m in the 5th second.Find the acceleration and the distance traveled in the 6th second
after 4 s, the velocity is 4*a
after 5 s, the velocity is 5*a
the average velocity in the 5th sec is
... (5a + 4a) / 2 = 4.5a
4.5a = 36 ... a = 8 m/s^2
To find the acceleration and the distance traveled, we can use the equations of motion.
1. From rest to the 5th second:
The particle covers a distance of 36m in the 5th second. We need to find the acceleration.
The equation that relates distance, acceleration, and time is:
s = ut + (1/2)at^2
Where:
s = distance
u = initial velocity (0, since the particle starts from rest)
t = time
a = acceleration
Plugging in the values:
36 = 0 + (1/2) * a * 5^2
36 = (1/2) * a * 25
a = (36 * 2) / 25
a ≈ 2.88 m/s^2
Therefore, the acceleration is approximately 2.88 m/s^2.
2. Distance traveled in the 6th second:
To find the distance traveled in the 6th second, we need to find the final velocity after 5 seconds and use it to calculate the distance traveled in the 6th second.
The equation that relates final velocity, initial velocity, acceleration, and time is:
v = u + at
Where:
v = final velocity
u = initial velocity (0, since the particle starts from rest)
a = acceleration (2.88 m/s^2, as calculated above)
t = time (5 seconds)
Plugging in the values:
v = 0 + 2.88 * 5
v = 14.4 m/s
Now, to find the distance covered in the 6th second, we use the equation:
Distance in 6th second = (v * time) + (1/2) * a * time^2
Where:
v = final velocity (14.4 m/s)
a = acceleration (2.88 m/s^2)
time = 1 second
Plugging in the values:
Distance in 6th second = (14.4 * 1) + (1/2) * 2.88 * 1^2
Distance in 6th second ≈ 14.4 + 1.44
Distance in 6th second ≈ 15.84 m
Therefore, the distance traveled in the 6th second is approximately 15.84 meters.