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The velocity and acceleration vectors of a particle undergoing circular motion are v=2i^m/s and a=2i^+4j^m/s2 respectively at an instant of time. The radius of the circle is
A. 1m
B. 2m
C. 3m
D. 4m

Answer
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Hint: You could write down the given velocity and acceleration into its x and y components. As the tangential velocity is found to be in the x direction, the centripetal acceleration which is always perpendicular to tangential velocity will be in the y direction. Now we could substitute these values in the expression for centripetal acceleration to get the required answer.
Formula used:
Expression for centripetal acceleration,
ac=v2r

Complete answer:
In the question, we are given the velocity and acceleration vectors of a particle as v=2i^ms1 and a=2i^+4j^ms2 respectively at an instant of time. We are asked to find the radius of the circle using these given information.
So, the velocity of the particle is given as,
v=2i^ms1
We could represent this velocity into its components as,
vx=2ms1
vy=0ms1
Acceleration is given as,
a=2i^+4j^ms2
So, we could represent this acceleration into its components as,
ax=2ms2
ay=4ms2
But we know that the tangential velocity contributes to the centripetal acceleration and also, the centripetal acceleration will be directed perpendicular to it towards the centre. As the velocity here is in the x-direction, the y component of the given acceleration will be the centripetal acceleration.
Now, let us recall the expression for centripetal acceleration which is given by,
ac=v2r
r=v2a
Substituting vx anday, we get,
r=224
r=1m
Therefore, we found the radius of the circle to be 1m.

Hence option A is the right answer.

Note:
When the movement of any body is found to be along the circumference of a circle, that body is said to be in circular motion. Motion of satellites around planets, motion of a stone tied in a thread, etc are some examples to it. As the direction of the velocity is seen to be changing constantly, the body will have a centripetal acceleration towards the centre of the circular path.
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