Question

a particle moves along the y - axis so that at time t ≥ 0 its position is given by y(t)=\frac{2}{3}t^{3}-5t^{2}+8t. over the time interval 0 < t < 5, for what values of t is the speed of the particle increasing?
(a) 2.5 < t < 5
(b) 4 < t < 5 only
(c) 0 < t < 1 and 4 < t < 5
(d) 1 < t < 2.5 and 4 < t < 5

Explanation:

Step1: Find the velocity function

The velocity $v(t)$ is the derivative of the position function $y(t)$. Given $y(t)=\frac{2}{3}t^{3}-5t^{2}+8t$, using the power - rule $\frac{d}{dt}(t^{n}) = nt^{n - 1}$, we have $v(t)=y^\prime(t)=2t^{2}-10t + 8$.

Step2: Find the acceleration function

The acceleration $a(t)$ is the derivative of the velocity function. So $a(t)=v^\prime(t)=4t-10$.

Step3: Find the critical points of the velocity function

Set $v(t)=2t^{2}-10t + 8 = 0$. Factor out a 2: $2(t^{2}-5t + 4)=0$, then $t^{2}-5t + 4=(t - 1)(t - 4)=0$. The critical points of $v(t)$ are $t = 1$ and $t = 4$.

Step4: Determine when speed is increasing

Speed is increasing when $v(t)$ and $a(t)$ have the same sign.

• First, find when $a(t)=4t - 10=0$, we get $t = 2.5$.
• Test intervals:
• For $0\lt t\lt1$: $v(0.5)=2(0.5)^{2}-10(0.5)+8=2\times0.25 - 5 + 8=0.5+3 = 3.5>0$, $a(0.5)=4\times0.5-10=2 - 10=-8<0$, speed is decreasing.
• For $1\lt t\lt2.5$: $v(2)=2(2)^{2}-10(2)+8=8 - 20 + 8=-4<0$, $a(2)=4\times2-10=-2<0$, speed is increasing.
• For $2.5\lt t\lt4$: $v(3)=2(3)^{2}-10(3)+8=18 - 30 + 8=-4<0$, $a(3)=4\times3-10=2>0$, speed is decreasing.
• For $4\lt t\lt5$: $v(4.5)=2(4.5)^{2}-10(4.5)+8=2\times20.25-45 + 8=40.5-45 + 8=3.5>0$, $a(4.5)=4\times4.5-10=8>0$, speed is increasing.

Answer:

D. $1\lt t\lt2.5$ and $4\lt t\lt5$