Question

A proton in a cyclotron changes its velocity from $30\, km/s$ north to $40 \,km/s$ east in $20s$. What is the magnitude of average acceleration during this time?

• A. $ 2.5\,km/{{s}^{2}} $
• B. $ 12.5\,km/{{s}^{2}} $
• C. $ 22.5\,km/{{s}^{2}} $
• D. $ 32.5\,km/{{s}^{2}} $

Answer 1

Answer: (A)

Change in velocity $=\sqrt{(40)^{2}+(30)^{2}} $
$=50\, km / s $

Average acceleration $=\frac{50}{20} $
$=2.5 \,km / s ^{2}$