Consider a particle moving along a straight line, whose position as a function of time is given by s(t)= t^2- t+ , where =1 ms ⁻² , =6 ms ⁻¹ and =5 m. The average speed of the particle, in ms ⁻¹ , from t=0 to t=6 s is :

Step-by-step solution

Given s(t) = t^2 - 6t + 5 Velocity v(t) = ds dt = 2t - 6 The particle comes to rest when v(t) = 0 , which gives t = 3 s. Position at t = 0 s is s(0) = 5 m. Position at t = 3 s is s(3) = 3^2 - 6(3) + 5 = -4 m. Position at t = 6 s is s(6) = 6^2 - 6(6) + 5 = 5 m. Distance travelled from t = 0 to t = 3 s is d₁ = |-4 - 5| = 9 m. Distance travelled from t = 3 to t = 6 s is d₂ = |5 - (-4)| = 9 m. Total distance travelled is D = d₁ + d₂ = 9 + 9 = 18 m. Average speed = 18 6 = 3 ms ⁻¹ Answer: 3