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A particle $P$ travels in a straight line. It passes through the point $O$ of the line with velocity $5{\text{ }}m{\text{ }}{s^{ - 1}}$ at time $t = 0$, where $t$ is in seconds. P’s velocity after leaving $O$ is given by

$\left( {0.002{t^3} - 0.12{t^2} + 1.8t + 5} \right){\text{ }}m{\text{ }}{s^{ - 1}}$

The velocity of $P$ is increasing when $0 \lt t \lt {T_1}$ and when $t \gt {T_2}$, and the velocity of $P$ is decreasing when ${T_1} \lt t \lt {T_2}$.

a) Find the values of ${T_1}$ and ${T_2}$ and the distance $OP$ when $t = {T_2}$.

b) Find the velocity of $P$ when $t = {T_2}$ and sketch the velocity-time graph for the motion of $P$.

پاسخ تشریحی :
نمایش پاسخ

a) $a\left( t \right) = 0.006{t^2} - 0.24t + 1.8$

$\left[ {0.006\left( {{t^2} - 40t + 300} \right) = 0} \right]$

${T_1} = 10$, ${T_2} = 30$

$s\left( t \right) = 0.0005{t^4} - 0.04{t^3} + 0.9{t^2} + 5t + \left( C \right)$

$\left[ {405 - 1080 + 810 + 150} \right]$

Distance is $285{\text{ }}m$

b) Velocity is $5{\text{ }}m{s^{ - 1}}$

For curve with $v$ increasing from $a + ve$

value at $t = 0$ to a maximum

Then decreases to $a + ve$ minimum and thereafter increases

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