Question

A car travels towards North for $10$ minutes with a velocity of $60kmph$, turns towards East and travels for $15$ minutes with a velocity of $80kmph$ and then turns towards North East and travels for $5$ minutes with a velocity of $60kmph$. Then what is its average acceleration (in $m/s^2$)?

• A. $7.08\times {10}^{-3}$
• B. $7.08\times {10}^{-2}$
• C. $1.8\times {10}^{-1}$
• D. $1.87\times {10}^{-3}$

Answer 1

The correct option is A $7.08\times {10}^{-3}$

Initial velocity is 60km/h in north direction.

Final velocity is 60km/h in north east direction ($30\sqrt{2}$ in north and $30\sqrt{2}$ in east.

Average acceleration is change in velocity in a given time period

Total time is 30min.

Change in velocity is $V_f-V_i$ $=>30\sqrt{2}east,(30\sqrt{2}-60)north$

Average acceleration $=60\sqrt{2}east,60(\sqrt{2}-2)north$

Magnitude of acceleration as it is a vector quantity.

$\sqrt{(60\sqrt{2}{)}^2+(60(\sqrt{2}-2){)}^2}$$=60\sqrt{8-4\sqrt{2}}$ in $km/h^2$

so for converting to $m/s^2$ multiply by 1000 and divide by $(3600{)}^2$ so acceleration is $=60\sqrt{8-4\sqrt{2}}\times \frac{1000}{{3600}^2}$ in $m/s^2$

Average acceleration is $7.08\times {10}^{-3}$ $m/s^2$.