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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
10 months ago

Answers : (2)

Arun
20882 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
 
 
Regards
Arun (askIITians forum expert)
10 months ago
Nikhil P
20 Points
							
We know acceleration, a =\frac{dv}{dt} =4t.
At t=1,a=4*1=4m/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
 
10 months ago
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