We can find the velocity (v) of the particle at any time by taking the derivative of its position (x) with respect to time (t):v = dx/dt
For the given position function, x = 2.5t^2, we have:
v = d(2.5t^2)/dt = 5t
Therefore, the velocity of the particle at time t is 5t m/s.
To find the velocity at t = 5 seconds, we substitute t = 5 into the expression for v:
v = 5t = 5(5) = 25 m/s
Hence, the speed of the particle at t = 5 seconds is 25 m/s. Note that speed is the magnitude of velocity and is always non-negative, so we don't need to include a sign.
Answer. C
A bead P sliding on a frictionless semi-circular string... bead Q ejected... relation between $t_P$ and $t_Q$ is 
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
Method used for separation of mixture of products (B and C) obtained in the following reaction is: 