Question

An object is travelling in a horizontal circle with uniform speed. At time =0 the velocity was 20i+35j and after one minute velocity become -20i-35j what is magnitude of a Accerlation

Answer 1

Answer:

300√65 m/s²

Explanation:

Initial velocity $=20\hat i+35\hat j$

Final velocity $=-20\hat i-35\hat j$

The final velocity is just equal and opposite

i.e. in 1 minute, the object has completed a half circle.

Displacement = (Final velocity - Initial velocity)/time

$=\frac{-20\hat i-35\hat j-(20\hat i+35\hat j)}{60}$

$=-\frac{40\hat i+70\hat j}{60}$

$=-\frac{4\hat i+7\hat j}{6}$

The magnitude of the displacement is equal to the twice the radius of the circle

Therefore, the radius of the circular path

$R=\frac{1}{2}\times\sqrt{(\frac{4}{6})^2+(\frac{7}{6})^2}$

$=\frac{1}{12}\sqrt{16+49}$

$=\frac{\sqrt{65}}{12}$

The magnitude of the velocity is

$V=\sqrt{(20)^2+(35)^2}$

$=\sqrt{400+1225}$

$=\sqrt{1625}$

The magnitude of the acceleration

$A=\frac{V^2}{R}$

$\implies A=\frac{1625}{\sqrt{65}/12}$

$\implies A=\frac{12\times 65\times 25}{\sqrt{65}}$

$\implies A=300\times \sqrt{65}$ m/s²

or, $A=300\sqrt{65}$ m/s²

Hope this helps.