A body starting from rest, travels a distance of 45 meters during the fifth second of motion.Calculate the acceleration of the body
v=u=at
d(6)-d(5)= 45
1/2 a (6^2-5^2)=45
surely you can solve for a.
To calculate the acceleration of the body, we need to use the kinematic equation for displacement:
\[
s = ut + \frac{1}{2}at^2
\]
Where:
- \(s\) is the displacement (45 meters in this case)
- \(u\) is the initial velocity (given as 0, as the body starts from rest)
- \(a\) is the acceleration (what we need to find)
- \(t\) is the time (given as 5 seconds)
Rearranging the equation to solve for acceleration:
\[
a = \frac{2(s - ut)}{t^2}
\]
Plugging in the values:
\[
a = \frac{2(45 - 0 \cdot 5)}{5^2} = \frac{90}{25} = 3.6 \, \text{m/s}^2
\]
Therefore, the acceleration of the body is \(3.6 \, \text{m/s}^2\).