The acceleration of a pARTICLE IS DIRECTLY PROPORTIONAL TO THE SQUARE OF THE TIME t. When t = 0, the particle is at x = 24 m. knowing that at t = 6 s, x = 96 m and v = 18 m/s, express x and v in terms of t.

Question

anonymous · Answer

The acceleration is directly proportional to the square of t, so in other words if k is an arbitrary constant:

$$\huge a = k*t^2 $$

Integrating with respect to t to get v:

$$\huge v = \int\limits\limits(a, t) = 1/3 * k * t^3 + C  $$

Where C is a constant of integration.

Integrating again to get an expression for the distance x:

$$\huge x = \int\limits\limits(v, t) = 1/12 * k * t^4 + C*t + D  $$

Where D is a second constant of integration.

Now you have two equations with three unknowns (eqns for v and x)

Use your initial conditions to solve for k, C, and D.

$$\huge 1) t=0, x=24 m $$

$$ \huge\ 24 = D \rightarrow D = 24 m  $$

$$\huge 2) t=6, x=96 m $$

$$96 = 1/12*k*(6^4) + C*6 + 24 \rightarrow C=-6 $$

$$\huge 3) t=6, v=18 $$

$$\huge 18 = 1/3*k*(6^(3)) - 6 \rightarrow  k=1/3   $$

anonymous · Answer

thank you..im waiting for the continuation of the answer..thank you so much

anonymous · Answer

$$\huge \mathcal{W}\mathcal{E}\mathcal{L}\mathcal{C}\mathcal{O}\mathcal{M}\mathcal{E}$$
$$\huge \mathbb{T}\mathbb{o}$$
$$\huge \mathit{OPENSTUDY}$$

anonymous · Answer

you may medal me if it helped :)

anonymous · Answer

thanks GAnesh

anonymous · Answer

@grudge what continuation? as i see it, @Rohangrr already gave the answer.

anonymous · Answer

lol thanks

anonymous · Answer

why did C became negative (-) 6?

anonymous · Answer

The answer is incorrect. When they first solved for C they forgot about the K variable. You first have to solve for C in terms of K then plug that back into the velocity equation to solve for K. Then solve for C. The answer should be K = 1/9 and C = 10.