Question

the position of a particle moving along x axis is given by x=t^2-5t+6where x is in m and t in second then speed of particla at origin is

1 .1 m per second
2. -1 m per second
3. 6 m per second​

Answer 1

Answer:

Given

$x = {t}^{2} - 5t + 6$

Differentiate wrt t

$\frac{dx}{dt} = v = 2t - 5$

At origin x = 0 , So

$x = 0 = {t}^{2} - 5t + 6 \\ \\ (t - 2)(t - 3) = 0 \\ \\ t = 2 \: and \: 3$

So

$v = 2(2) - 5 = - 1 \: m {s}^{ - 1} \\ and \\ v = 2(3) - 5 = 1 \: m {s}^{ - 1}$

So particle passes through origin twice at t=2 and 3 secs .

Ans is 1 and 2