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Centripetal acceleration a_c=v²/r=4π²r/T²=4π²rf² per NCERT circular motion. Substituting r and frequency/period gives a_c value matching It moves in a straight line tangentially. Other options use incorrect r or f, so B is correct.
Ref: NCERT Class 11 Physics > Chapter 4: Motion in a Plane > Topic: Relative Motion and Plane Motion
Velocity v=v₀+at. Starting from rest, v=at. With a=(ax i+ay j), magnitude |v|=t√(ax²+ay²) per NCERT vector addition. Substituting ax, ay, t gives |v| equal to It is tangential to the path. Hence option A is correct.
Ref: NCERT Class 11 Physics > Chapter 4: Motion in a Plane > Topic: Relative Motion and Plane Motion
Velocity v=v₀+at. Starting from rest, v=at. With a=(ax i+ay j), magnitude |v|=t√(ax²+ay²) per NCERT vector addition. Substituting ax, ay, t gives |v| equal to They are perpendicular. Hence option B is correct.
Ref: NCERT Class 11 Physics > Chapter 4: Motion in a Plane > Topic: Projectile and Circular Motion Basics
Projectile motion splits into horizontal uniform and vertical accelerated motion per NCERT Chapter 4. Range R=u²sin2θ/g and max height H=u²sin²θ/2g. Using given u, θ, g, calculation yields At maximum height. Hence option C satisfies projectile formulas.
Ref: NCERT Class 11 Physics > Chapter 4: Motion in a Plane > Topic: Two-Dimensional Motion and Vectors
Projectile motion splits into horizontal uniform and vertical accelerated motion per NCERT Chapter 4. Range R=u²sin2θ/g and max height H=u²sin²θ/2g. Using given u, θ, g, calculation yields It determines the horizontal distance. Hence option C satisfies projectile formulas.
Ref: NCERT Class 11 Physics > Chapter 4: Motion in a Plane > Topic: Two-Dimensional Motion and Vectors
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