Given a particle moving in the positive direction has an acceleration $a=(100-4v^2)m/s^2$

To find the time interval and displacement of a particle when speed changes from $1 m/s$ to $3 m/s.$

• Let x be the displacement in meters v be velocity and a be the acceleration
• We know that velocity$v=\dfrac{\mathrm dx}{\mathrm dt}$ so$x=\int v \ \mathrm dt$
• Acceleration$a={\dfrac{\mathrm dv} {\mathrm dt}}={\dfrac{\mathrm dv}{ \mathrm dx} }{\dfrac{\mathrm dx}{ \mathrm dt}}=v\dfrac{\mathrm dv}{\mathrm dx}$ so$\int \mathrm dt=\int \dfrac{\mathrm dv}{a}$and$\mathrm dx=\dfrac{v \mathrm dv}{ a}=\dfrac{v}{100-4v^2}\mathrm dv \cdots(1)$
• Time$t=\int {\dfrac{1} {100-4v^2}\mathrm dv}=\int \dfrac{1} {10^2-(2v)^2}\mathrm dv$$=\dfrac{1}{20} \int {\bigg(\dfrac{1}{10+2v}+\dfrac{1}{10-2v}}\bigg)\ \mathrm dv$$=\dfrac{1}{40}\bigg(\ln(10+2v)-\ln(10-2v)\bigg)+C$(we know that$\dfrac{\mathrm d(\ln(x))}{\mathrm dt}=\dfrac{1}{ x}\mathrm dx $ and$\int \dfrac{1}{ x}\mathrm dx=\ln(x)+C$ * So time$t= \dfrac{1}{ 40}{\ln\bigg(\dfrac{10+2v}{10-2v}\bigg)}+C$ * Displacement$x=\int vdt$ * From (1) we get * So$x=\int \dfrac{v}{100-4v^2}dv=-\dfrac{1}{8}\ln(100-4v^2)+C$ * Time interval=$t=t(v=2m/s)-t(v=1m/s)=\dfrac{1}{40}\bigg(\ln\bigg(\dfrac{10+2(2)}{10-2(2)}\bigg)-\ln\bigg(\dfrac{10+2(1)}{10-2(1)}\bigg)\bigg)=0.011046s$ * Displacement=$x = -\dfrac{1}{8}(\ln(100-16)-\ln(100-4))=0.0167m$ * So the time interval and displacement of a particle when speed changes from$1 m/s$ to$3 m/s$ are 0.011046s and 0.0167m respectively