Answer:
V = (√2) + 1) m/s
Explanation:
Let the mass and speed of the running student be M and V respectively.
We are told that when he speeds up by 1 m/s, he has the same kinetic energy as his brother.
Thus, his speed at which he mow has the same kinetic energy as his brother is (V + 1) m/s
Now, we are told that the mass of the student is twice as large as that of his brother. Thus, his brother's mass is; M/2
Since kinetic energy is given by the formula K.E = ½mv²
Therefore, since we want to find the original speed of both students and that the initial condition says that the running student had half the kinetic energy of the brother, we now initial condition as;
½MV²= ½(½(M/2)V²) - - - - (eq 1)
Since he has sped up by 1 m/s, and has a kinetic energy now equal to that of his brother, we have;
(½M(V + 1)²) = (½(M/2)V²) - - - - (Eq2)
Dividing eq 1 by eq 2 gives;
V²/(V + 1)²= 1/2
Taking square root of both sides gives;
V/(V + 1) = 1/√2
Cross multiply to give;
(√2)V = V + 1
(√2)V - V = 1
V((√2) - 1) = 1
V = 1/((√2) - 1)
Simplifying this using surfs gives;
V = [1/((√2) - 1)] × ((√2) + 1))/((√2) + 1))
V = ((√2) + 1))/1
V = (√2) + 1) m/s
