Answer to Question #128851 in Mechanics | Relativity for Ariyo Emmanuel

Question

The position of a particle is given by S = (2t^2 − 8t + 6)m, where t is in seconds. Determine the time when the velocity of the particle is zero, and the total distance travelled by the particle when t = 3 s.

Answer 1

$s(t) = 2t^2-8t+6$ m. The velocity is $v = \frac{dS}{dt}= 2 \cdot 2t -8 = 4t - 8$ m/s.

$v=0 \Rightarrow 4t-8 = 0 \Rightarrow t=2$ s.

So, when t = 2s the velocity is equal to zero. To find the total distance travelled by the particle by t = 3s, we need to take into account that particle at first was moving with positive speed, then it stopped and changed direction of motion into opposite, and after that was moving till the position s(3).

$s(0) = 0 - 0+6 = 6$ m (start of motion).

$s(2) = 2 \cdot4 - 8 \cdot2+6 = -2$ m (turnoff point).

$s(3) = 2 \cdot9 - 8 \cdot3 + 6 = 18 - 24 + 6 =0$ m (end of motion).

In total we have 2 parts of the path: from 6m to -2m, and from -2m to 0m.

The total distance is $s_T= 6 +2+2 = 10$ m.

Answer: $v = 0$ m/s when $t=2$ s; $s_T = 10$ m.