Physics - Kinematics Question with Solution | TestHub

$\mathrm{a}=24-200{\mathrm{t}}^2=\frac{\mathrm{dv}}{\mathrm{dt}}$

$={\int }_0^{\mathrm{v}}\mathrm{dv}={\int }_0^{\mathrm{t}}24-200{\mathrm{t}}^2\mathrm{dt}$

$\mathrm{v}=24\mathrm{t}-\frac{200{\mathrm{t}}^3}{3}$

for $0\le \mathrm{t}\le 0.2$ at $\mathrm{t}=0.1$,

$$\mathrm{v}=\frac{7}{3}\mathrm{m}/\mathrm{s}$$

$\frac{\mathrm{dx}}{\mathrm{dt}}=24\mathrm{t}-\frac{200{\mathrm{t}}^3}{3}$

${\int }_{0.8}^{\mathrm{x}}\mathrm{dx}={\int }_{0.4}^{0.2}24\mathrm{t}-\frac{200{\mathrm{t}}^3}{3}\mathrm{dt}$

$\mathrm{x}-0.8={\left(\frac{24{\mathrm{t}}^2}{2}-\frac{50{\mathrm{t}}^4}{3}\right)}_{0.4}^{0.2}$

$\mathrm{x}=-\mathrm{ve}$

for $\mathrm{t}=0.3\sec$

at $\mathrm{t}=0.2\sec ,\mathrm{v}=\frac{128}{30}$

for $\mathrm{t}=0.2\sec$ to 0.3 sec

Area of curve $=\frac{1}{2}(16+8)\times 0.1=1.2$

$\Rightarrow \mathrm{v}-\frac{128}{30}=1.2$

at $\mathrm{t}=0.3\sec ,\mathrm{v}=\frac{128+36}{30}=\frac{164}{30}$