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.
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
In the given circuit the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 Ω, the value of the current in the circuit will be A.
For $ \alpha, \beta, \gamma \in \mathbb{R} $, if $$ \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2} - 3}{\sin 2x - \beta x} = 3, $$ then $ \beta + \gamma - \alpha $ is equal to: