question a particle travels along the x - axi...
question a particle travels along the x - axis such that its velocity is given by v(t)=(t^1.9 - 2)cos(3t). what is the acceleration of the particle at time t = 3? you may use a calculator and round your answer to the nearest thousandth. answer attempt 2 out of 2 additional solution no solution submit answer
Answer
# Explanation:
## Step1: Recall acceleration - velocity relationship
Acceleration $a(t)$ is the derivative of velocity $v(t)$. So we need to find $v^\prime(t)$ using the product - rule. The product - rule states that if $y = u(t)v(t)$, then $y^\prime=u^\prime(t)v(t)+u(t)v^\prime(t)$. Let $u(t)=t^{1.9}-2$ and $v(t)=\cos(3t)$.
## Step2: Find $u^\prime(t)$ and $v^\prime(t)$
Using the power - rule, if $y = t^n$, then $y^\prime=nt^{n - 1}$. So, $u^\prime(t)=1.9t^{0.9}$. Using the chain - rule, if $y=\cos(ax)$, then $y^\prime=-a\sin(ax)$. So, $v^\prime(t)=- 3\sin(3t)$.
## Step3: Apply the product - rule
$a(t)=v^\prime(t)=(1.9t^{0.9})\cos(3t)+(t^{1.9}-2)(-3\sin(3t))$.
## Step4: Evaluate $a(t)$ at $t = 3$
$a(3)=(1.9\times3^{0.9})\cos(9)+(3^{1.9}-2)(-3\sin(9))$.
Using a calculator:
$3^{0.9}\approx2.62074$, $1.9\times3^{0.9}\approx1.9\times2.62074 = 4.9794$. $\cos(9)\approx0.9111$, so $(1.9\times3^{0.9})\cos(9)\approx4.9794\times0.9111\approx4.537$.
$3^{1.9}\approx6.968$, $3^{1.9}-2\approx4.968$, $\sin(9)\approx - 0.412$, so $(3^{1.9}-2)(-3\sin(9))\approx4.968\times(-3)\times(-0.412)\approx6.147$.
$a(3)\approx4.537 + 6.147=10.684$.
# Answer:
$10.684$