a body moving in a straight line with an acceleration of 12ms-1 has a velocity of 8ms-1 after 5 seconds. find the initial velocity of the body and the distance covered in the 9th second
v(t) = v0 + at
so plug in your numbers at solve for v0.
The distance covered is
s(t) = v0*t + 1/2 at^2
the distance covered in the 9th second is
s(9) - s(8)
so now just plug and chug.
Please elaborate on your answer
To find the initial velocity and the distance covered in the 9th second, we need to use the equations of motion.
1. The formula to calculate the final velocity (vf) when the initial velocity (vi), acceleration (a), and time (t) are known is:
vf = vi + (a * t)
2. The formula to calculate the distance covered (s) when the initial velocity (vi), acceleration (a), and time (t) are known is:
s = (vi * t) + (0.5 * a * t^2)
Let's solve for the initial velocity (vi) first:
Given:
Final velocity (vf) = 8 m/s
Acceleration (a) = 12 m/s^2
Time (t) = 5 s
Using equation (1):
8 = vi + (12 * 5)
8 = vi + 60
vi = 8 - 60
vi = -52 m/s
The initial velocity is -52 m/s, which means the body was moving in the opposite direction of its final velocity.
Now, let's find the distance covered in the 9th second:
The time at the end of the 8th second is 8 seconds. So, we need to calculate the distance covered between 8 seconds and 9 seconds.
Given:
Initial velocity (vi) = -52 m/s
Acceleration (a) = 12 m/s^2
Time (t) = 9 s (time at the end of the 9th second) - 8 s (time at the end of the 8th second) = 1 s
Using equation (2):
s = (vi * t) + (0.5 * a * t^2)
s = (-52 * 1) + (0.5 * 12 * 1^2)
s = -52 + 6
s = -46 meters
The distance covered in the 9th second is -46 meters, indicating that the body moved 46 meters in the opposite direction of its initial motion.