A particle moved along the x axis with a constant acceleration of -85.0 meters per second squared. The initial position of the particle was at the origin and its initial velocity was +325 meters per second. Find the position of the particle at T=4.0 seconds.
Please walk through your steps,
Thank you very much.
v=vₒ-at.
Assume v=0, than
t= vₒ/a =325/85=3.82 s.
It takes 3.82 s. to stop, so after 4 s the particle
will be at rest at the point
x=vₒt-at²/2 =325•3.82 - 85•(3.82)²/2 =621.3 m
Thank you so much!!
To find the position of the particle at T=4.0 seconds, you can use the following kinematic equation:
x = xo + vo * t + (1/2) * a * t^2
Where:
x = final position
xo = initial position
vo = initial velocity
t = time
a = acceleration
Plugging in the given values:
xo = 0 (particle was initially at the origin)
vo = +325 m/s
a = -85.0 m/s^2 (acceleration is negative because it is acting in the opposite direction of the positive x-axis)
t = 4.0 s
Substituting these values into the equation, we get:
x = 0 + (325 m/s) * (4.0 s) + (1/2) * (-85.0 m/s^2) * (4.0 s)^2
First, let's calculate the value of (4.0 s)^2:
(4.0 s)^2 = 16 s^2
Now, let's substitute this back into the equation:
x = 0 + (325 m/s) * (4.0 s) + (1/2) * (-85.0 m/s^2) * (16 s^2)
x = 1300 m - 680 m
x = 620 m
Therefore, the position of the particle at T=4.0 seconds is 620 meters.