The figure shows the velocity and acceleration of a point like body at the intial moment of its motion. The acceleration vector of the body remains constant. The minimum radius of the curvature of trajectory of the body is :
Details and Assumptions
v ∘ = 8 m s − 1
a = 2 m s − 1
θ = 1 5 0 ∘
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A level 1 question.
v= v0-root 3 /2 at
after resolving a in parallel and perpendicular components of v , we get acos60 = v^2/r solving it we get r=8
Can you please explain what you did?
حلّك مصنج، حيث حلّي أفضل
Wrong answer bro
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Did you solve it that you are saying it is wrong, if did ,upload!
What did you just do?
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As The Acceleration Vector Remains Constant So Its Magnitude Will Obviously Remain Constant.
We Know that
Radius Of Curvature = v*v/a
where "a" is the component of total acceleration perpendicular to the direction of instantaneous velocity.
For R To Be minimum the component of "a" perpendicular to velocity should be maximum.That Can Happen
When The Whole Acceleration Vector Becomes Perpendicular To Velocity. Hence We Need To Find The Time
When Velocity Vector Will Be Perpendicular To Acceleration Vector .
So Simply We Can Write The Velocity Vector At Any instant of time t using the first equation of kinematics in
vector form (v=u+at). then we can take dot product of velocity vector and acceleration vector and put it
equal to zero to get the required time . Now we can backsubstitute this time into the expression for velocity
vector to get the magnitude of velocity at that instant.