Question

Racing greyhounds are capable of rounding corners at very high speeds. A typical greyhound track has turns that are 45m diameter semicircles. A greyhound can run around these turns at a constant speed of 16m/s .part a: What is its acceleration in m/s2?partb: What is its acceleration in units of g ?

Answer 1

To solve this problem we will apply the concept related to centripetal acceleration. Normal acceleration or centripetal acceleration is responsible for changing the direction of the velocity vector. It is the only type of acceleration present in the uniform circular motion. Its mathematical formula is given by

$a = \frac{v^2}{r }$

Here,

v = Tangential velocity

r = Radius

Our values are,

$v = 16m/s$

$r = \frac{45m}{2} = 22.5$

PART A) Using the previous expression the acceleration will be

$a= \frac{16^2}{22.5 }$

$a= 11.37 m/s^2$

PART B) In the case of g units, we know that

$9.8m/s^2 = 1g$

Then performing the conversion we have to

$a = 11.37m/s^2(\frac{1g}{9.8m/s^2})$

$a = 1.16g$