Question

The acceleration of an athlete at 54 kmph when negotiting a turn of radius 45m is

Answer 1

Answer:

Given:

Velocity while turning = 54 km/hr

Radius of turn = 45 m

To find:

Acceleration of the athlete

Concept:

While undergoing a turn , an object always experiences a force which remains directed radially inwards . It is directed towards the centre of the circular track in which the turn is being taken.

The acceleration is dependent on the velocity of the object and radius of track.

The acceleration is called Centripetal Acceleration.

Conversion:

Speed has to be converted to m/s unit.

54 km/hr = 15 m/s

Calculation:

let the acceleration be denoted as "a".

$\boxed{ \huge{ \blue{a = \dfrac{ {v}^{2} }{r} }}}$

$\implies \: a = \dfrac{ {(15)}^{2} }{45}$

$\implies \: a = 5 \: m {s}^{ - 2}$

So final answer :

$\boxed{ \huge{ \red{ \bold{\: a = 5 \: m {s}^{ - 2} }}}}$

Answer 2

AnswEr :

Explanation :

From the Question,

• Velocity of the athlete (v) = 54 Km/h

• Radius of the path (r) = 45 m

Firstly,

We would have to convert Km/h into m/s

Multiply by 5/18

Thus,54 Km/h = 15 m/s

NoTE

While executing circular motion,a body would be accelerating along both the radius and tangent

• Radial Acceleration
• Tangential Acceleration

Here,

The tangential acceleration is under consideration

Thus,

$\boxed{\boxed{\sf a_c = \dfrac{v^2}{r}}}$

(Putting the values)

$\longrightarrow \sf a_c = \dfrac{15 \times 15}{45} \\ \\ \longrightarrow \sf a_c = \dfrac{15}{3} \\ \\ \longrightarrow \sf a_c = 5 m/s^2$

Hence,acceleration of the athlete is 5 m/s²