A particle start moving with acceleration 2m/s^2.Distance covered by it in 5th half sec is?
at the end of the 4th half sec, the velocity is 4 m/s
at the end of the 5th half sec, the velocity is 5 m/s
the average velocity over the 5th half second is 4.5 m/s
To find the distance covered by the particle in the 5th half-second, we can use the equation of motion:
S = ut + (1/2)at^2
Where:
S = distance covered
u = initial velocity (assumed to be zero since it wasn't given)
a = acceleration (given as 2 m/s^2)
t = time in seconds (given as 0.5 seconds)
Now, let's calculate the distance covered in the 5th half-second.
First, we need to calculate the time elapsed for the 5th half-second.
To calculate the time for the 5th half-second, we can use the formula:
t = n * 0.5
Where:
n = the number of the half-second (in this case, 5)
Plugging in the values, we have:
t = 5 * 0.5
t = 2.5 seconds
Now that we have the time, we can substitute the values into the equation:
S = (0 * 2.5) + (1/2) * 2 * (2.5)^2
Simplifying further:
S = 0 + (1/2) * 2 * (2.5)^2
S = (1/2) * 2 * (6.25)
S = 6.25 meters
Therefore, the distance covered by the particle in the 5th half-second is 6.25 meters.