Question

A car moving towards north at a speed of 10 m/s
turns right within 10 s maintaining the same speed.
Average acceleration of the car during this time
interval is
(1) 5 m/s2 S-W
(3) V2 m/s2 S-E
(2) 5 m/s? N-E
(4) V2 m/s2 N-E​

Answer 1

Answer:

root 2 m/s NE

Explanation:

Initial Velocity due to north (u) = 10 m/s

Final velocity east (v) = 10m/s

Time interval = 10 sec

=> root 10 square + 10 square = root 200 = 10 root 2 m/s (change in velocity)

=> accelaration (a) = v-u/t

=> 10 root 2 / 10

=> root 2 m/s

Answer 2

Explanation:

Given that,

Initial speed, u = 10 m/s (in north)

Final speed, v = 10 m/s (in east)

Due to the change in direction, a acceleration will produced by the car. Resultant velocity,

$v'=\sqrt{10^2+10^2}\ =10\sqrt{2}\ m/s$

Time taken, t = 10 s

So, average acceleration is given by :

$a=\dfrac{v-u}{t}$

$a=\dfrac{10\sqrt{2}}{10}$

$a=\sqrt{2}\ m/s^2$

So, the acceleration of the car is $a=\sqrt{2}\ m/s^2$. Hence, this is the required solution.