A truck travelling with uniform acceleration crosses two points A B with velocities 60 m / s and 40 m / s respectively. The speed of the body at the mid point of A and B is nearest to:

Question

A truck travelling with uniform acceleration crosses two points A B with velocities 60 m / s and 40 m / s respectively. The speed of the body at the mid point of A and B is nearest to: (A) 17 m / s (B) 20 m / s (C) 19.49 m / s (D) 50.9 m / s

Tardigrade Admin · Accepted Answer

As we know the relation V text mid =√(VA2+VB2/2) V mid =√((60)2+(40)2/2)=√(3600+1600/2) V text mid =√(5200/2)=√2600=50.9 m / s

Q. A truck travelling with uniform acceleration crosses two points $A & B$ with velocities $60 m / s$ and $40 m / s$ respectively. The speed of the body at the mid point of $A$ and $B$ is nearest to:

17 m / s

20 m / s

19.49 m / s

50.9 m / s

As we know the relation $V_{mid} = \frac{V _{A}^{2} + V _{B}^{2}}{2}$
$V_{mi d} = \frac{( 60 ) ^{2} + ( 40 ) ^{2}}{2} = \frac{3600 + 1600}{2}$
$V_{mid} = \frac{5200}{2} = 2600 = 50.9 m / s$

Q. A truck travelling with uniform acceleration crosses two points A&B with velocities 60m/s and 40m/s respectively. The speed of the body at the mid point of A and B is nearest to: