A particle starts from the origin with a velocity of 10 m/s and moves with a constant acceleration till the velocity increases to 50 m/s. At that instant the acceleration is suddenly reversed. What will be the velocity of the particle, when it returns to the starting point?
avg velocity =30m/s
vf^2=vi^2+2ad
2500=100+2a d or d=1200/a
So new issue: we apply -a to stop it first.
vf=vi+at=50-a*t or t=50/a to stop.
it now has went a second distance of
vf^2=Vi^2+2ad
0=50^2-2a d or
d=1250/a+1200/a
now on the return trip from the max total distance to the original position
vf^2=vi^2+2ad
vf^2=2(-a)(-1250/a-1200/a(
vf^2=2*2450
vf=sqrt(4900( =70m/s in the opposite direction of the original 10m/s
check my math.
To find the velocity of the particle when it returns to the starting point, we can break the problem into two parts: the first part when the particle is accelerating and the second part when the particle is decelerating.
First, let's find the time it takes for the particle to reach its maximum velocity. We can use the equation of motion:
v = u + at,
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken. Let's denote the time taken for the particle to reach its maximum velocity as t1.
Given:
Initial velocity, u = 10 m/s
Final velocity, v = 50 m/s
We can rearrange the equation to solve for t1:
t1 = (v - u) / a,
Substituting the given values into the equation:
t1 = (50 - 10) / a,
Now, let's find the distance traveled by the particle during the first part. We can use the equation:
s = ut + (1/2)at^2,
where s is the distance traveled.
Since the particle starts from the origin, the distance traveled in the first part will be:
s1 = (1/2)at1^2,
Now, let's find the time it takes for the particle to return to the starting point after the acceleration is reversed. Let's denote this time as t2.
During this time, the particle will decelerate with the same magnitude of acceleration. Therefore, the time taken to return to the starting point will be the same as the time taken to reach the maximum velocity, t1:
t2 = t1,
Now, let's find the distance traveled by the particle during the second part, which can be obtained using the equation:
s2 = v*t2 - (1/2)a*t2^2,
Since the particle returns to the starting point, the total distance traveled will be the sum of s1 and s2:
Total distance traveled (s) = s1 + s2,
Using the values of s1 and s2, we can find the total distance traveled by the particle.
Finally, to find the velocity of the particle when it returns to the starting point, we can use the equation of motion:
v = u + at,
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken. Let's denote the final velocity as v2.
Given:
Initial velocity, u = 10 m/s
Final velocity, v2 = ?
Acceleration, a = -a (since it is reversed)
Rearranging the equation, we have:
v2 = u - a*t2 = u - a*t1,
Substituting the values of u, a, and t1 into the equation, we can find the final velocity of the particle when it returns to the starting point.