Question

sladerThe position s(t) of a particle moving along the x-axis at time t is given by s(t) = −t3 + 3t2 + 5 4 , where s is measured in meters and t is measured in seconds. At what time is the particle's instantaneous velocity equal to its average velocity on the interval [0, 6]?

Answer 1

Answer:

t = 3.646 seconds

Step-by-step explanation:

We have given instantaneous velocity

$s(t) = -t^3 + 3t^2 + 5 4$

Now,

$\dfrac {ds}{dt} = -3t^2 + 6t ...(1)$

Now we will find the average velocity

$\text {Average velocity} =\dfrac {s(t_2) - s(t_1)}{t_2 - t_1}$

$\text {Average velocity} = \dfrac {s(6) - s(0)}{6 - 0} = \dfrac{((-6)^3 + 3\times 6^2 + 54)- (0 + 3\times 0 + 54)}{6} \\\\\text {Average velocity} = \dfrac {-216 + 108 + 54 - 54}6 = \dfrac {-108}6 =-18 ... (2)\\\\ \text {Average velocity} = -18$

It is also given that particle's instantaneous velocity equal to its average velocity.

From equation (1) and (2)

$-3t^2 + 6t = -18\\3t^2 - 6t - 18 = 0\\t^2 -2t - 6 = 0$

Now we factorize the equation by the rule of Shri dharacharya

$\dfrac {-b \pm \sqrt{b^2 - 4ac} }{2a}$

$t=\dfrac {-2 \pm\sqrt{(-2)^2 - 4 \times 1 \times -6} }{2 \times 1} \\\\t=\dfrac {-2 \pm\sqrt{ 4 + 24} }{2 } \\\\t= \dfrac {-2 \pm\sqrt{ 28} }{2 }\\\\t= \dfrac {-2 \pm 2\sqrt{ 7} }{2 }\\t= 1 \pm \sqrt 7$

If we put the value of $\sqrt{7}$

If we take positive value than

t = 1 + 2.646 = 3.646

If we take negative value than

t = 1 - 2.646 =  -1.646

We can not take negative value of time So, t = 3.646 seconds

Answer 2

The instantaneous velocity of the particle equals the average velocity at $t \approx 3.646 \,s$.

How to find times in a kinematic function by mean value theorem

Given a curve differentiable at $[a, b]$, there is a value $c \in [a, b]$ such that:

$f'(c) = \frac{f(b)-f(a)}{b-a}$ (1)

Where $f'(c)$ is the first derivative of the function evaluated at $c$.

Now we find the ends of the secant line and the expression for the first derivative:

$s(0) = -0^{3} + 3\cdot (0)^{2}+\frac{5}{4}$

$s(0) = \frac{5}{4}$

$s(6) = -(6)^{3}+3\cdot (6)^{2}+\frac{5}{4}$

$s(6) = -\frac{427}{4}$

$f'(c) = -3\cdot c^{2}+6\cdot c$

Then, we apply the mean value theorem:

$-3\cdot c^{2}+6\cdot c = \frac{-\frac{427}{4}-\frac{5}{4} }{6-0}$

$-3\cdot c^{2}+6\cdot c +18 =0$

The roots of this second order polynomial are: $c_{1}\approx 3.646$, $c_{2} \approx -1.646$.

Only the former one is inside the given interval. Hence, the instantaneous velocity of the particle equals the average velocity at $t \approx 3.646 \,s$. $\blacksquare$

To learn more on mean value theorem, we kindly invite to check this verified question: https://brainly.com/question/13950067