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 timev,
interval is​

Answer 1

As the car turns, acceleration is produced.

So, initial velocity due north (u)=10 m/s

Final velocity east (v)=10 m/s

Time interval (t)=10 s

Following vectors:- √10²+10²=√200=10√2 m/s=change in velocity

So, acceleration (a)=v-u/t

=10√2/10

=√2 m/s

So, average acceleration is equal to √2 m/s