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.
The correct Answer is 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: $g = 10\,\text{m s}^{-2}$) 
Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).
