Question

the acceleration of an athlete at 54ķmph when negotiating a turn of radius 45m​

Answer 1

Speed of athlete = 54*5/18 = 15m/s

Acceleration = v^2 /R = 15*15/45 = 5m/s^2

Explanation:

Answer 2

Given:

The speed of of an athlete (v) = 54 $\dfrac{km}{hr}$ and

The radius of an athlete (r) = 45 m

We have to find, the acceleration of an athlete (a) = ?

Solution:

The speed of of an athlete (v) = 54 $\dfrac{km}{hr}$

= 54 $\dfrac{1000m}{3600s}$ [∵ 1 km = 1000 m and 1 hr = 3600 s]

= 54 $\times \dfrac{5m}{18s}$ = 15 $\dfrac{m}{s}$

We know that,

The acceleration of an athlete (a) = $\dfrac{v^2}{r}$

= $\dfrac{15^2}{45} \dfrac{m}{s^2}$ = $\dfrac{15\times 15}{45} \dfrac{m}{s^2}$

= 5 $\dfrac{m}{s^2}$

∴ The acceleration of an athlete (a) = 5 $\dfrac{m}{s^2}$

Thus, the acceleration of an athlete (a) is 5 $\dfrac{m}{s^2}$.