The position of an object as a function of time is given as x = At3 + Bt2 + Ct + D. The constants are A = 2.10m/s3, B = 1m/s2, C = −4.10m/s, and
D = 3m.(a) at what time(s) is the object at rest
at rest means that the velocity dx/dt = 0
so, you need to solve
3At^2 + 2Bt + C = 0
now plug in your numbers and work your magic.
To find the time(s) at which the object is at rest, we need to determine when the velocity of the object is zero. The velocity can be obtained by taking the derivative of the position function with respect to time.
Given x = At^3 + Bt^2 + Ct + D, let's find the velocity function by calculating its derivative:
v = dx/dt
v = 3At^2 + 2Bt + C
To find the time(s) at which the object is at rest, we set the velocity function equal to zero and solve for t:
3At^2 + 2Bt + C = 0
Substituting the values of A, B, and C, we have:
3(2.10m/s^3)t^2 + 2(1m/s^2)t + (-4.10m/s) = 0
Now we can solve this quadratic equation for t using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
Where a = 3(2.10m/s^3), b = 2(1m/s^2), and c = -4.10m/s
Plugging in the values, we have:
t = (-(2m/s^2) ± √((2m/s^2)^2 - 4(3m/s^3)(-4.10m/s))) / (2(3m/s^3))
Simplifying further:
t = (-2m/s^2 ± √(4m^2/s^4 + 49.2m^2/s^2)) / (6m/s^3)
t = (-2m/s^2 ± √(53.2m^2/s^2)) / (6m/s^3)
t = (-2m/s^2 ± √(53.2m^2/s^2)) / (6m/s^3)
Now, we solve for t:
t = (-2m/s^2 ± √(53.2m^2/s^2)) / (6m/s^3)
Using a calculator, we can solve for t:
t ≈ -0.899s or t ≈ 0.278s
Therefore, the object is at rest at approximately -0.899 seconds and 0.278 seconds.
To find the time(s) at which the object is at rest, we need to determine when the velocity of the object is equal to zero. In other words, we need to find the values of 't' for which the derivative of position with respect to time is zero.
The derivative of the position function with respect to time is the velocity function. So, let's find the derivative:
v = dx/dt = d(A*t^3 + B*t^2 + C*t + D)/dt
Differentiating each term separately:
v = 3A*t^2 + 2B*t + C
Setting this expression equal to zero and solving for 't':
3A*t^2 + 2B*t + C = 0
Substituting the given values for A, B, and C:
3(2.10m/s^3)*t^2 + 2(1m/s^2)*t + (-4.10m/s) = 0
6.30*t^2 + 2*t - 4.10 = 0
Now, we can solve this quadratic equation to find the values of 't' when the object is at rest.
Using the quadratic formula,
t = (-b ± √(b^2 - 4ac)) / (2a)
where a = 6.30, b = 2, and c = -4.10.
Substituting these values into the formula:
t = (-2 ± √(2^2 - 4*6.30*(-4.10))) / (2*6.30)
Simplifying further:
t = (-2 ± √(4 + 103.2)) / 12.6
t = (-2 ± √107.2) / 12.6
Now, calculate the values of 't' using a calculator or software:
t ≈ -0.184 seconds or t ≈ 0.484 seconds
Therefore, the object is at rest at approximately t = -0.184 seconds and t = 0.484 seconds.