A particle starts from rest and moves with a constant acceleration of 0.5m/s2. Calculate the time taken by the particle to cover a distance of 25m.
vf^2=vi^2+2ad
vf^2=2*.5*25=25
vf=5m/s
time= distance/avgvelocity=25/(5/2)=10 seconds
To calculate the time taken by the particle to cover a distance of 25m, we can use the equation of motion:
\[s = ut + \frac{1}{2}at^2\]
Where:
s = distance traveled (25m in this case)
u = initial velocity (0 m/s, as the particle starts from rest)
a = acceleration (0.5 m/s^2, given)
t = time taken to cover the distance
Using the equation, we can rearrange it to solve for t:
\[t = \sqrt{\frac{2s}{a}}\]
Plugging in the values:
\[t = \sqrt{\frac{2 \times 25}{0.5}}\]
Simplifying:
\[t = \sqrt{\frac{50}{0.5}}\]
\[t = \sqrt{100}\]
\[t = 10\]
Therefore, the time taken by the particle to cover a distance of 25m is 10 seconds.
To calculate the time taken by the particle to cover a distance of 25m, we can use the kinematic equation:
s = u * t + (1/2) * a * t^2
Where:
s is the distance covered,
u is the initial velocity (which is 0 as the particle starts from rest),
a is the constant acceleration, and
t is the time taken.
In this case, s = 25m and a = 0.5m/s². Since the particle starts from rest, u = 0.
We can rearrange the equation to solve for t:
s = (1/2) * a * t^2
Multiply both sides of the equation by 2:
2s = a * t^2
Divide both sides of the equation by a:
t^2 = (2s) / a
Take the square root of both sides:
t = √[(2s) / a]
Let's substitute the given values into the equation:
t = √[(2 * 25m) / 0.5m/s²]
t = √[50m / 0.5m/s²]
t = √[100s²]
t = 10s
Therefore, the time taken by the particle to cover a distance of 25m is 10 seconds.