The position of an object as a function of time is given as x = At3 + Bt2 + Ct + D. The constants are A = 2.3m/s3, B = 1.8m/s2, C = −4.3m/s, and
D = 2 m.(a) What is the velocity of the object at t = 12s?
(b) What is the acceleration of the object at t = 0.5s?
well I did the one you posted later
This time you take the derivative
x = A t^3 + B t^2 + C t + D
v = 3 A t^2 + 2 B t + C
a = 6 A t + 2 B
v = dx/dt = 3A t^2 + 2B t + C
solve for t=12
v = dx/dt = 3A t^2 + 2B t + C
solve for t=12
a = dv/dt = 6A t + 2B
solve for t=0.5
To find the velocity of the object at a specific time, we need to take the derivative of the position function with respect to time. This will give us the rate of change of position, which is velocity.
(a) Velocity of the object at t = 12s:
To find the velocity of the object at t = 12s, we first need to find the derivative of the position function.
Given: x = At^3 + Bt^2 + Ct + D
Take the derivative of x with respect to t:
dx/dt = 3At^2 + 2Bt + C
Now plug in the values of the constants:
A = 2.3 m/s^3
B = 1.8 m/s^2
C = -4.3 m/s
dx/dt = 3(2.3)(12^2) + 2(1.8)(12) - 4.3
Evaluate the expression:
dx/dt = 3(2.3)(144) + 2(1.8)(12) - 4.3
= 993.6 + 43.2 - 4.3
= 1032.5 m/s
Therefore, the velocity of the object at t = 12s is 1032.5 m/s.
(b) Acceleration of the object at t = 0.5s:
To find the acceleration of the object at t = 0.5s, we need to find the derivative of the velocity function, which is the derivative of the position function.
Take the second derivative of x with respect to t:
d^2x/dt^2 = d/dt(3At^2 + 2Bt + C)
Now, take the derivative of the velocity function:
d^2x/dt^2 = d(3At^2 + 2Bt + C)/dt
= 6At + 2B
Plug in the values of the constants:
A = 2.3 m/s^3
B = 1.8 m/s^2
d^2x/dt^2 = 6(2.3)(0.5) + 2(1.8)
= 6.9 + 3.6
= 10.5 m/s^2
Therefore, the acceleration of the object at t = 0.5s is 10.5 m/s^2.