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
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 \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to: