An object moves across a straight line so that after t seconds, is distance from its starting point is: D(t)=t^3 - 12t^2 + 100t + 12 meters. Find the acceleration of the object after 3 seconds.
I believe acceleration is calculated by taking the second derivative.
D'(t) = 3t^2 - 24t + 100
D"(t) = 6t - 24
D"(3) = 18 - 24
D"(3) = -6
Hmmm...I have my doubts because it's a negative number.
You did everything correct, D"(t) is the acceleration. It happens to be decelerating at t=3 and at t=4 the acclereation turns positive.
Observe that the velocity is still positive at t=3.
Oh, I see! So even though the acceleration is negative at t=3, the object is still moving forward because its velocity is positive. It's like when you're driving a car and you hit the brakes, the car slows down (negative acceleration) but is still moving forward. I guess that's why they call it deceleration! Thanks for pointing that out!
If the velocity of the object is positive at t=3, it means that the object is still moving in the positive direction at that particular moment. However, the negative acceleration implies that the object is slowing down.
In simpler terms, even though the object is moving in the positive direction at t=3, it is slowing down. This is because the magnitude of the acceleration is decreasing.
So, the object is decelerating at 3 seconds, meaning it is slowing down while still moving in the positive direction.
You are correct in your calculation of the acceleration using the second derivative. The acceleration is given by the formula: D''(t) = 6t - 24.
To find the acceleration after 3 seconds, substitute t = 3 into the equation:
D''(3) = 6(3) - 24 = 18 - 24 = -6.
The negative sign indicates that the object is decelerating at this time. Note that the negative acceleration does not imply a negative velocity. In this case, the velocity is still positive, meaning the object is moving in the positive direction along the straight line. The negative acceleration simply indicates that its speed is decreasing.
It is important to mention that at t = 4, the acceleration will become positive, indicating that the object starts to accelerate again.