(12) A particle is moving in a straight line. Its position $x$ at time $t$ is given by
$x = 18 - 24 t + 9 t^{2} - t^{3} 0 ⩽ t ⩽ 5$
(i) Find the velocity $v$ at time $t$ and the values of $t$ for which $v = 0$ .
(ii) Find the position of the particle at those times.
(iii) Find the total distance travelled in the interval $0 ⩽ t ⩽ 5$ .

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Answer

Finally, the total distance travelled in the interval $0 ⩽ t ⩽ 5$ is $28$ units.

Steps

Step 1 ：First, we find the velocity function by taking the derivative of the position function with respect to time: $v (t) = - 3 t^{2} + 18 t - 24$ .

Step 2 ：Next, we find the values of $t$ for which $v = 0$ . We get $t = 2$ and $t = 4$ .

Step 3 ：Then, we find the position of the particle at those times: $x (2) = - 2$ and $x (4) = 2$ .

Step 4 ：To find the total distance travelled, we find the position at $t = 0$ and $t = 5$ , and then sum the absolute differences between consecutive positions: $x (0) = 18$ and $x (5) = - 2$ .

Step 5 ：Finally, the total distance travelled in the interval $0 ⩽ t ⩽ 5$ is $28$ units.

(12) A particle is moving in a straight line. Its position x at time t is given byx=18−24t+9t2−t30⩽t⩽5(i) Find the velocity v at time t and the values of t for which v=0.(ii) Find the position of the particle at those times.(iii) Find the total distance travelled in the interval 0⩽t⩽5.

Answer

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