A body accelerates uniformly from rest and covers a distance of 180m in 6s. Calculate the magnitude of its acceleration.
. solution
S=ut+1/2at^2
180=0x6+1/2xax6^2
180=0+a/2x6^2
180-0=36a/2
180=36a/2
180x2=36a
360=36a
360/36=36a/36
10ms^-2=a
acceleration is 10ms^-2
looks good, but I'd have gone from
180 = 36a/2
to
180 = 18a
Why did the body accelerate so quickly? Because it had a need for speed, of course! The magnitude of its acceleration is 10 m/s². That's one speedy body!
To find the magnitude of the acceleration, we can use the formula S = ut + (1/2)at^2, where S is the distance covered, u is the initial velocity (which is 0 since the body starts from rest), a is the acceleration, and t is the time elapsed.
Given:
S = 180m
u = 0
t = 6s
Plugging in these values into the formula, we get:
180m = 0(6s) + (1/2)a(6s)^2
Simplifying:
180m = 0 + 3a(36s^2)
180m = 108as^2
Dividing both sides by 108s^2:
180m / (108s^2) = as^2
Simplifying:
5/3 = a
Therefore, the magnitude of the acceleration is 5/3 m/s^2 or approximately 1.67 m/s^2.
To calculate the magnitude of the acceleration, we can use the equation of motion S = ut + (1/2)at^2, where S is the distance covered, u is the initial velocity (which is zero in this case since it started from rest), a is the acceleration, and t is the time taken.
Plugging in the given values:
S = 180m
u = 0
t = 6s
The equation becomes:
180 = 0(6) + (1/2)a(6)^2
180 = 0 + (1/2)a(36)
180 = 18a
To find the value of a, we divide both sides of the equation by 18:
180/18 = 18a/18
10 = a
Therefore, the magnitude of the acceleration is 10 m/s^2.