3:For an object whose velocity in ft/sec is given by v(t) = −2t^2 + 4, what is its distance travelled, in feet, on the interval t = 0 to t = 2 secs?
a:4.876 b:2.667 c:9.752 d:2.438
the answer is 4.876
I took the exam and put 2.667 and got it wrong so here's the right answer
then the distance formula is
s = (-2/3)t^3 + 4t + c, where c is a constant
at t = 0, s = c
at t = 2, s = -16/3 + 8 + c
distance in the given interval
= -16/3 + 8 + c - c
= 8/3 which is close to choice b)
^^^thats wrong
To find the distance traveled by the object on the interval t = 0 to t = 2 seconds, you need to integrate the velocity function over that interval.
First, let's rewrite the velocity function v(t) as a function of distance x. The velocity is the rate of change of distance with respect to time, so we can express it as:
v(t) = dx/dt
Now, we can rearrange this equation to solve for dx:
dx = v(t) * dt
Integrating both sides of the equation gives the distance traveled:
∫dx = ∫v(t) * dt
The interval of integration is from t = 0 to t = 2.
∫[0 to 2] dx = ∫[0 to 2] (−2t^2 + 4) * dt
Now, we can compute the integral:
∫[0 to 2] (−2t^2 + 4) * dt = ∫[0 to 2] (−2t^2) dt + ∫[0 to 2] 4 dt
Evaluating each integral:
= [-2/3 * t^3] from 0 to 2 + [4t] from 0 to 2
Plugging in the values and evaluating:
= [-2/3 * (2)^3 - (-2/3 * (0)^3)] + [4 * 2 - 4 * 0]
Simplifying:
= [-2/3 * 8] + [8 - 0]
= -16/3 + 8
= -16/3 + 24/3
= 8/3
So, the distance traveled by the object on the interval t = 0 to t = 2 seconds is 8/3 feet.
However, none of the answer choices provided match this result. Please double-check the given options or the question statement.