- Home
- Standard 11
- Physics
3-2.Motion in Plane
hard
A particle starts from rest and performing circular motion of constant radius with speed given by $v = \alpha \sqrt x$ where $\alpha$ is a constant and $x$ is the distance covered. The correct graph of magnitude of its tangential acceleration $(a_t)$ and centripetal acceleration $(a_c)$ versus $t$ will be:
A
B
C
D
Solution
$\mathrm{v}=\alpha \sqrt{\mathrm{x}} \Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=\alpha \sqrt{\mathrm{x}} \Rightarrow \mathrm{x}=\frac{\alpha^{2} \mathrm{t}^{2}}{4}$$a_{t}=v \frac{d v}{d x}=\alpha \sqrt{x} \cdot \frac{\alpha}{2 \sqrt{x}}=\frac{\alpha^{2}}{2}=$ constant
$a_{c}=\frac{v^{2}}{r}=\frac{\alpha^{2} x}{r}=\frac{\alpha^{2}}{r} \times \frac{\alpha^{2} t^{2}}{4}$
$a_{c} \propto t^{2}$
Standard 11
Physics
