Question

A plane flies 10 km east from city A to city B in 2 hrs and then 20 km south from city B to city C in 1 hr. For the total trip, what are the magnitude and direction of its average acceleration?

• A. $\frac{-5}{3}\hat{i}-\frac{20}{3}\hat{j}$
• B. $\frac{5}{3}\hat{i}-\frac{20}{3}\hat{j}$
• C. $South-eastdirection$
• D. $South-westdirection$

Answer 1

Answer: (A), (D)

$South-westdirection$

Average acceleration $=\frac{{\overrightarrow{v}}_2-{\overrightarrow{v}}_1}{t}$

$\overrightarrow{v_1}=\frac{10}{2}\hat{i}=5\hat{i}$ (East direction)

$\overrightarrow{v_2}=\frac{20}{1}\left(\widehat{-j}\right)=-20\hat{j}$ (south direction)

$\therefore {\overrightarrow{a}}_{avg}=\frac{-20\hat{j}-5\hat{i}}{3}$(time of flight=3hrs)

$=\frac{-20}{3}\hat{j}-\frac{5}{3}\hat{i}$

since, both x component and y component are negative, it is in the south west direction.