The position function x(t) of a particle moving along an x axis is x = 6.00 - 8.00t2, with x in meters and t in seconds. (a) At what time and (b) where does the particle (momentarily) stop? At what (c) negative time and (d) positive time does the particle pass through the origin?
Take the derivative to get v = -16t then find where v = 0 which is at t=0 in this case therefore
A = 0
To find where is momentarily stops you must input 0 into the original function of x there we end up with 6 -8(0)^2 and this means the position of t = 0 is where x = 6
B = 6
To find the negative time and positive time solve for t given that x = 0, since x is equal to 6 - 8t^2 subtract 6 from both for
-6 = -8t^2
Then divide by -8 for
.75 = t^2
then take the square root of both sides for
.86602 ~ t
when time is negative just multiply both sides by -1
-.86602 ~ -t
Therefore
C ~ -.86602
D ~ .86602
To find the answers to the given questions, we need to analyze the position function in relation to time.
(a) To find the time at which the particle momentarily stops, we need to find the velocity function and set it equal to zero. The velocity function v(t) is the derivative of the position function x(t).
Taking the derivative of x(t):
v(t) = d(x)/dt = d(6.00 - 8.00t^2)/dt = -16.00t
Setting v(t) equal to zero:
-16.00t = 0
t = 0
So, the particle momentarily stops at t = 0 seconds.
(b) To find the position at which the particle momentarily stops, we substitute the time value we found (t = 0) into the position function x(t):
x(0) = 6.00 - 8.00(0)^2 = 6.00 meters
Therefore, the particle momentarily stops at x = 6.00 meters.
(c) To find the negative time at which the particle passes through the origin, we need to find the time when the position function is equal to zero (x = 0). We can solve this equation:
0 = 6.00 - 8.00t^2
Rearranging the equation:
8.00t^2 = 6.00
Dividing both sides by 8.00:
t^2 = 0.75
Taking the square root of both sides (considering the negative root for a negative time):
t = -√0.75 ≈ -0.866
So, the particle passes through the origin at a negative time of t ≈ -0.866 seconds.
(d) Similarly, to find the positive time at which the particle passes through the origin, we can consider the positive root of the equation:
t = √0.75 ≈ 0.866
Therefore, the particle passes through the origin at a positive time of t ≈ 0.866 seconds.
To solve this problem, we need to find the time and position at which the particle stops, as well as the negative and positive times at which it passes through the origin.
(a) To find the time at which the particle stops, we need to determine when the velocity of the particle is zero. The velocity is the derivative of the position function with respect to time. The derivative of x(t) = 6.00 - 8.00t^2 is:
v(t) = dx/dt = -16.00t
Setting v(t) = 0 and solving for t:
-16.00t = 0
t = 0
Therefore, the particle stops at t = 0 seconds.
(b) To find the position at which the particle stops, we substitute the value of t = 0 into the position function:
x(0) = 6.00 - 8.00(0)^2
x(0) = 6.00
Therefore, the particle stops at x = 6.00 meters.
(c) To find the negative time at which the particle passes through the origin, we need to solve the position function for x = 0:
0 = 6.00 - 8.00t^2
Rearranging the equation:
8.00t^2 = 6.00
Dividing both sides by 8:
t^2 = 0.75
Taking the square root:
t = ±√(0.75)
Since we're only interested in the negative time, we take the negative value:
t ≈ -0.87 seconds
Therefore, the particle passes through the origin at t ≈ -0.87 seconds.
(d) To find the positive time at which the particle passes through the origin, we take the positive square root:
t ≈ +√(0.75)
t ≈ +0.87 seconds
Therefore, the particle passes through the origin at t ≈ +0.87 seconds.