Question

A particle moves along the positive x-axiz with an acceleration $a_x$ in meter per meter per second squared which increases linearly with x expressed in centimeter, as shown on the graph for an interval of its motion if the velocity of the particle at $x=4cm$ is $0.40m/s$ determine the velocity at $x=12cm$

• A. $0.2m/s$
• B. $0.4m/s$
• C. $0.8m/s$
• D. $1.6m/s$

Answer 1

The correct option is C $0.8m/s$

Area under curve $=\int adx=\int vdv=\frac{v_t^2}{2}-\frac{v_1^2}{2}$

$\Rightarrow \frac{1}{2}\times 6\times \frac{8}{100}=\frac{v_1^2}{2}-\frac{0.4\times 0.4}{2}\to 0.48=v_t^2-0.16$

$\Rightarrow V_t^2=0.64\Rightarrow v_r=0.8m/s$