A particle residing in one dimension is moving with constant acceleration a=-4m/sec^2 .At time t =2sec, the particle is located at x=3m and is observed to have velocity v=6m/sec. (a) At what instants is the particle to be found at x=0m?(ii)what is the velocity of the particle at each of these instant? (b)(i)At what instant is the particle stopped ,i.e., v=0m/sec?(ii) where is the particle
a = -4
v = -4t+c; v(2)=6, so c=14
v = -4t+14
s = -2t^2+14t+c; s(2) = 3, so c = -17
s = -2t^2 + 14t - 17
Now you can answer the questions
A particle residing in one dimension is moving with constant acceleration a=-4m/sec^2 .At time t =2sec, the particle is located at x=3m and is observed to have velocity v=6m/sec. (a) At what instants is the particle to be found at x=0m?(ii)what is the velocity of the particle at each of these instant? (b)(i)At what instant is the particle stopped ,i.e., v=0m/sec?(ii) where is the particle
To solve these problems, we'll use the equations of motion for constant acceleration:
1. Position: x = x_0 + v_0t + (1/2)at^2
2. Velocity: v = v_0 + at
a) To find the instant when the particle is at x = 0 m:
1. Substitute the known values into the position equation:
0 = 3 + 6t + (1/2)(-4)t^2
2. Rearrange the equation to obtain a quadratic equation:
-2t^2 + 6t + 3 = 0
3. Solve the equation using any appropriate method (factoring, quadratic formula, etc.). In this case, using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = -2, b = 6, c = 3:
t = (-6 ± √(6^2 - 4*(-2)*3)) / (2*(-2))
t = (-6 ± √(36 + 24)) / (-4)
t = (-6 ± √60) / -4
Simplifying further:
t = (-6 ± √(4 * 15)) / -4
t = (-6 ± 2√15) / -4
Dividing by -2:
t = 3/2 ± (√15) / 2
This gives us two possible values for t.
b) i) To find when the particle is stopped (v = 0 m/s):
1. Substitute the known values into the velocity equation:
0 = 6 + (-4)t
2. Rearrange the equation to solve for t:
4t = 6
t = 6/4
t = 3/2
This gives us the instant when the particle's velocity is 0 m/s.
ii) To find where the particle is at that instant, substitute this value of t back into the position equation:
x = 3 + 6(3/2) + (1/2)(-4)(3/2)^2
x = 3 + 9 - 9/4
x = 27/4
Therefore, the particle is at x = 27/4 m at the instant when its velocity is 0 m/s.