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.
The motion of an airplane is represented by the velocity-time graph as shown below. The distance covered by the airplane in the first 30.5 seconds is km.

For \( \alpha, \beta, \gamma \in \mathbb{R} \), if \[ \lim_{x \to 0} \frac{x^2 \sin(\alpha x) + (\gamma - 1)e^{x^2}}{\sin(2x - \beta x)} = 3, \] then \( \beta + \gamma - \alpha \) is equal to: