Question

For an object whose velocity in ft/sec is given by v(t) = −t2 + 2, what is its displacement, in feet, on the interval t = 0 to t = 2 secs?
1.333
2.438
2.667
−4

Answer 1

Answer:

1.333

Step-by-step explanation:

We know that the displacement function is the integral of the velocity function. We are also given our bounds, 0 seconds to 2 seconds.

x(t) = $\int\limits^2_0 {t^{2} +2} \, dt$ = -t³/3 + 2t (evaluated from zero to two)

= [-2³/3 + 2(2)] - [-0³/3 + 2(0)] = 4/3 = 1.333...

Answer 2

Answer:

1.333

Step-by-step explanation:

Distance travelled (or displacement) will be the antiderivative of Velocity.

Antiderivative of -t^2 + 2 =

d(t) = -t^3/3 +2t

Now plug in t=2 and t=0 and subtract them.

(-8/3+4)- (0/3 +0)= -8/3+4 = 1.333