Question

thin 10 s maintaining the same speed.
A car moving towards north at a speed of 10 m/s
turns right within 10 s maintaining the s
Average acceleration of the car during this time
interval is
(1) 5 m/s2 S-W
N
(2) 5 m/s2 N-E
(3) V2 m/s2 S-E
(4) v2 m/s2 N-E
​

Answer 1

Explanation:

average acceleration = final velocity - initial velocity / time

$10 \div 10 = 1ms - {}^{2}$

Answer 2

The acceleration of the car during this interval is $\sqrt{2}\ m/s^2$ in north-east direction.

Explanation:

We have, a car moving towards north at a speed of 10 m/s  turns right within 10 s maintaining the same speed. When it takes a turn, it will reach to east direction.

The change in velocity of the car is given by :

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

The change in velocity per unit time is called acceleration of an object. So,

$a=\dfrac{10\sqrt{2}\ m/s}{10\ s}\\\\a=\sqrt{2}\ m/s^2$

The direction of car is in north-east direction.

so, the acceleration of the car during this interval is $\sqrt{2}\ m/s^2$ in north-east direction.

Learn more,

Acceleration

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