Question

For an object whose velocity in ft/sec is given by v(t) = −t^2 + 4, what is its displacement, in feet, on the interval t = 0 to t = 3 secs?

7.67
−3.00
−0.33
3.00

Answer 1

3 feet

Explanation:

The relationships between displacement (position), velocity and acceleration are:

`\boxed{\boxed{\begin{array}{c}\textbf{DISPLACEMENT (s)}\\\\\text{Differentiate} \downarrow\qquad\uparrow\text{Integrate}\\\\\textbf{VELOCITY (v)}\\\\\text{Differentiate}\downarrow\qquad\uparrow \text{Integrate}\\\\\textbf{ACCELERATION (a)}\end{array}}}`

Displacement is an object’s distance from a particular point (often its starting point), measured in a straight line. It’s not necessarily the same as the total distance travelled.

To find the displacement of an object in its first 3 seconds of travel, we need to integrate its velocity function on the interval [0, 3] seconds.

`\begin{aligned}\displaystyle \int^3_0 (-t^2+4)\; \text{d}t&=\left[(-t^(2+1))/(2+1)+4t\right]^3_0\\\\&=\left[-(t^(3))/(3)+4t\right]^3_0\\\\&=\left(-((3)^(3))/(3)+4(3)\right)-\left(-((0)^(3))/(3)+4(0)\right)\\\\&=-9+12+0-0\\\\&=3\end{aligned}`

Therefore, the displacement of the object on the interval [0, 3] seconds is 3 feet.

Answer 2

The answer is 3 feet

Explanation:

The displacement function is the integration of the velocity formula, so:

`d(t)=\int\limits^0_t \, (-t^(2) +4) dt \\d(t)=\int\limits^0_t{} \, (-t^(2)) dt + \int\limits^0_t{} \, 4 dt \\\\d(t)=((-t^(3) )/(3) + C ) + (4t + C)\\\\\\\ d(t)=frac{-t^(3) }{3} + 4t + C\\`

if we evaluate the displacement for the upper and lower limit it would be

D = total displacement = d(3) - d(0)

`D=((-3^(3)  )/(3)  + 12 + C) - ((-0^(3) )/(3)  + 4*0 + C)\\\\D=(-9  + 12 + C) - (0  + 0 + C)\\\\\\D=(3 + C) -  C\\\\\\D=3`