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Motion in A Plane question

2011 · Q148
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Motion in A Plane question

2011 · Q148

NEETPhysicsMotion in A PlaneMCQ+4 / −1
A body is moving with velocity 30 m/s towards east . After 10 seconds its velocity becomes 40 m/s towards north. The average acceleration of the body is
  1. A
    1 m/s2
  2. B
    7 m/s2
  3. C
    7\sqrt 77​ m/s2
  4. D
    5 m/s2
View written solutionFree

Correct answer: D

AIPMT 2011 Prelims Physics - Motion in a Plane Question 36 English Explanation

Velocity towards east direction, v1→=30i^\overrightarrow {{v_1}} = 30\widehat iv1​​=30i m/s

Velocity towards north direction, v2→=40j^\overrightarrow {{v_2}} = 40\widehat jv2​​=40j​ m/s

Change in velocity, Δv→=v→2−v→1\Delta \overrightarrow v = {\overrightarrow v _2} - {\overrightarrow v _1}Δv=v2​−v1​ = (40j^−30i^)\left( {40\widehat j - 30\widehat i} \right)(40j​−30i)

∴∣Δv→∣=∣40j^−30i^∣=50 m/s \therefore \left| {\Delta \overrightarrow v } \right| = \left| {40\widehat j - 30\widehat i} \right| = 50\,m/s∴​Δv​=​40j​−30i​=50m/s

Average acceleration, a→av{\overrightarrow a _{av}}aav​ = change in velocityTime interval{{change\,in\,velocity} \over {Time\, interval}}Timeintervalchangeinvelocity​

a→av=v→2−v→1Δt=Δv→Δt{\overrightarrow a _{av}} = {{{{\overrightarrow v }_2} - {{\overrightarrow v }_1}} \over {\Delta t}} = {{\Delta \overrightarrow v } \over {\Delta t}}aav​=Δtv2​−v1​​=ΔtΔv​

∣a→av∣=∣Δv→∣Δt=50 m/s10 s=5 m/s2\left| {{{\overrightarrow a }_{av}}} \right| = {{\left| {\Delta \overrightarrow v } \right|} \over {\Delta t}} = {{50\,m/s} \over {10\,s}} = 5\,m/{s^2}​aav​​=Δt∣Δv∣​=10s50m/s​=5m/s2

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