Question

For an object whose velocity in ft/sec is given by v(t) = -3t2 + 5, what is its displacement, in feet, on
the interval t = 0 to t = 2 secs?

Answer 1

Step-by-step explanation:

To find the displacement I'll give you two ways since I don't know if you had experience with calculus. So here's both ways.

Way 1: Graph 1: Geometric Apporoach.

I'll a graph where t is x axis and v is y axis. So we have a velocity versus time graph.

Then, we graph the function

Look Above this what graph I get

Now, to find displacement,

Remeber displacement is the product of velocity and time. So here displacement is area under the curve.

We want displacement from 0 to 2. so let form the shape under the curve.

Look at the second graph, we form a triangle and we get this

Area of Triangle is 1/2 bh.

The base is 2 and the height is -6 so we get

$\frac{1}{2} (2)( - 12) = - 12$

Our height is -12 because velocity is a vector so it can be negative and down is negative.

So the displacement is -12

Way 2: Calculus Way.

We can use integration to find the answer.

$∫(f(x)dx$

So now we have

$∫( - 3 {t}^{2} + 5)dx$

$∫( - 6t)$

Now we do,

$∫( - 6(0 ) + - 6(2) = - 12$

So the area under the curve is -12. Thus displacement is -12