Q15 Q15 A car travelling at 36 km/hr due North tums West in 5 s under constant acceleration - Science - Electricity

Question

Q15

Q15 A car travelling at 36 km/hr due North tums West in 5 s under
constant acceleration maintains the same speed The acceleration of
the car is

(A) 2 √ 2 m/s2 South-West (B) 10 m/s2
North-East

(C) 2 √ 2 m/s2 South East (D) 5 m/s2
North-West

Pintu B. · Accepted Answer

Dear Student ,
Initial velocity of car=36km/h=10m/s in north direction=10j

Final velocity of car=10 m/s in west direction=-10i

where i and j are the unit vectors along x axis and y axis respectively.

Change in velocity=-10i-10j

Resultant

$\sqrt{10^{2} + 10^{2}}$

$= 10 \sqrt{2}$

Acceleration =Change in velocity/Time taken

= $\frac{10 \sqrt{2}}{5} = 2 \sqrt{2} = 2.828 m / s^{2}$
Regards

Q15 Q15. A car travelling at 36 km/hr due North tums West in 5 s under constant acceleration maintains the same speed. The acceleration of the car is (A) 2 √ 2 m/s2 South-West (B) 10 m/s2 North-East (C) 2 √ 2 m/s2 South.East (D) 5 m/s2 North-West