Question

A constant force $F$ acts on a body of mass $1kg$ moving in the direction of motion with velocity $v_0$ for 1 second. The distance moved by the body is:

• A. $v_0-\frac{F}{2}$
• B. $v_0+\frac{F}{2}$
• C. $\frac{v_{0}F}{2}$
• D. ${2v}_{0}F$

Answer 1

Answer: (B)

$v_0+\frac{F}{2}$

Here, $F=ma=a$
so the acceleration of the body $=a=F$
Using , $S=ut+\frac{1}{2}a{t}^2$
here, $u=v_0,t=1s,a=F$
thus, $S=v_0+\frac{1}{2}F$