Physics - Kinematics Question with Solution | TestHub

Speed is greatest at x = 1 m. The area under the acceleration-displacement graph is a triangle.

$$v\frac{dv}{dx}=a(x)$$

$$\frac{1}{2}{v}_{\max }^2={\int }_0^{1}a(x)dx$$

$${\int }_0^{1}a(x)dx=\frac{1}{2}(1)(1)$$

$${\int }_0^{1}a(x)dx=\frac{1}{2}{\mathrm{m}}^2/{\mathrm{s}}^2$$

$$v_{\max }^2=1{\mathrm{m}}^2/{\mathrm{s}}^2$$

$$v_{\max }=1m/s$$

Option A.