Question

question 12 (1 point) the acceleration of a particle in meters per second per second varies with time according to the following function: $a(t)=12t^{2}-4$. at t = 2s the position of the particle is 6 m and the velocity is 22 m/s. the displacement of the particle from 4 to 5 seconds is a 309 m/s b 329 m/s c 349 m/s d 369 m/s e 389 m/s

Explanation:

Step1: Recall the relationship between acceleration, velocity and displacement

Velocity $v(t)$ is the integral of acceleration $a(t)$. Given $a(t)=12t^{2}-4$, then $v(t)=\int(12t^{2}-4)dt = 12\times\frac{t^{3}}{3}-4t + C=4t^{3}-4t + C$.
We know that at $t = 2s$, $v(2)=22m/s$. Substitute $t = 2$ into $v(t)$:
$v(2)=4\times2^{3}-4\times2 + C=22$.
$4\times8-8 + C=22$.
$32 - 8+C=22$.
$24 + C=22$, so $C=- 2$.
So $v(t)=4t^{3}-4t - 2$.

Step2: Recall the relationship between velocity and displacement

Displacement $s(t)$ from $t_1$ to $t_2$ is given by $s=\int_{t_1}^{t_2}v(t)dt$. We want to find the displacement from $t = 4s$ to $t = 5s$.
$s=\int_{4}^{5}(4t^{3}-4t - 2)dt$.
First, find the antiderivative of $4t^{3}-4t - 2$, which is $F(t)=4\times\frac{t^{4}}{4}-4\times\frac{t^{2}}{2}-2t=t^{4}-2t^{2}-2t$.
Then $s=F(5)-F(4)$.
$F(5)=5^{4}-2\times5^{2}-2\times5=625-50 - 10=565$.
$F(4)=4^{4}-2\times4^{2}-2\times4=256-32 - 8=216$.
$s=565 - 216=349m$.

Answer:

c. 349 m/s