The speed of a particle changes from 5 ~ms ⁻¹ to 2 5 ~ms ⁻¹ in a time t . If the magnitude of change in its velocity is 5 ~ms ⁻¹ , the angle between the initial and final velocities of the particle is

Step-by-step solution

Given, v_i= 5 ~ms ⁻¹, v_f=2 5 ~ms ⁻¹ and v=5 ~ms ⁻¹ Since, both v_i and v_f are extreme speeds, i.e. at t=0 and t=t . So, they can be considered as magnitude of the velocities at time, t=0 and t=t . As we know that R^2=A^2+B^2+2 A B Hence, the angle between the velocities, = v^2-v_i^2-v_f^2 2 v_i v_f Putting the given values, we get aligned & = (5)^2-( 5 )^2-(2 5 )^2 2( 5 )( 5 ) & = 25-5-20 10 = 0 10 =0 =90^ aligned Hence, the correct option is (d).