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A particle is performing circular motion of radius 1 m it's speed is v = 2 t^2 ( twice of time square ) What will its magnitude of acceleration at t= 1


2 years ago

Arun
24742 Points
							Dear student V = 2 t² Now Acceleration a = dV/dt a = 2 * 2t = 4t At t = 1 sec a = 4 m/sec² Hope it helps  RegardsArun (askIITians forum expert)

2 years ago
Nikhil P
20 Points
							We know acceleration, a =$\frac{dv}{dt}$ =4t.At t=1,a=4*1=4$m/s^{2}$  If the magnitude of tangential acceleration is the question,then as tangential acceleration and centripetal acceleration are perpendicular, $a^2=\sqrt{a_c^2+{a_t}^2}$$a^{2}={a_{t}}^{2}+{a_{c}}^{2}$${a_{t}}^{2}=a^{2}-{a_{c}}^{2}$${a_t}=\sqrt{a^{2}-{a_c}^{2}}}$ We know,   ${a_c}=\frac{v^{2}}{r}$ v,at t=1,=2*1*1=2 $\Rightarrow {a_c}=\frac{4}{1}$               =4 $\Rightarrow {a_c}=\sqrt{a^2-{a_t}^2} =\sqrt{16-4}=2\sqrt{3} m/s^2$

2 years ago
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