A body moves with a speed of 10 m/s in the curved path of 25 m radius of curvature. If the tangential acceleration is 3 m/s2, then total acceleration for the body will be:

  1. 3.3 m/s2
  2. 4 m/s2
  3. 5 m/s2
  4. None of these

Answer (Detailed Solution Below)

Option 3 : 5 m/s2
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Detailed Solution

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CONCEPT:

Centripetal Acceleration (ac): 

  • Acceleration acting on the object undergoing uniform circular motion is called centripetal acceleration.
  • It always acts on the object along the radius towards the center of the circular path.
  • The magnitude of centripetal acceleration,

\(a = \frac{{{v^2}}}{r}\)

Where v = velocity of the object and r = radius

Tangential acceleration (at):

  • It acts along the tangent to the circular path in the plane of the circular path.
  • Mathematically Tangential acceleration is written as

\(\overrightarrow {{a_t}} = \vec \alpha \times \vec r \)

Where α = angular acceleration and r = radius

CALCULATION:

Given – v = 10 m/s, r = 25 m and at = 3 m/s2

  • Net acceleration is the resultant acceleration of centripetal acceleration and tangential acceleration i.e.,

\(a = \sqrt {a_c^2 + a_t^2} \)

Centripetal Acceleration (ac):

\(\therefore {a_c} = \frac{{{v^2}}}{r}\)

\( \Rightarrow {a_c} = \frac{{{{\left( {10} \right)}^2}}}{{25}} = \frac{{100}}{{25}} = 4\;m/{s^2}\)

Hence, net acceleration 

\(a = \sqrt {a_t^2 + a_c^2} = \sqrt {{4^2} + {3^2}} = 5\;m/{s^2}\)
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