Question

An object is travelling in a horizontal circle with uniform speed. At0.the velocity is
given by u=20i+35ikm/s. After one minute the velocity becomes v=-20i -35j. what is the
magnitude f the acceleration?​

Answer 1

Answer:

Acceleration will be 1.666j m/sec^2

Explanation:

We have given initial velocity of the object u = 20i+35j km/sec

And final velocity is given as v = 20i -35j km/sec

Time is given as 1 minute

We know that 1 minute = 60 sec

Now according to third equation of motion

We know that v = u+at

So acceleration $a=\frac{v-u}{t}=\frac{(20i+35j)-(20i-35j)}{60}=1.666jm/sec^2$

Answer 2

Answer:

$|a|=1.41\ m/s^2$

Explanation:

It is given that,

Initial speed of the object, $u=(20i+35j)\ m/s$

Final speed of the object, $v=(-20i-35j)\ m/s$

Time taken, t = 1 min = 60 s

Let a is the acceleration of the object. It can be calculated as :

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

$a=\dfrac{(-20i-35j)-(20i+35j)}{60}$

$a=\dfrac{-40i-75j}{60}$

$a=(-0.666i-1.25j)\ m/s^2$

$|a|=\sqrt{(-0.666)^2+(-1.25)^2}$

$|a|=1.41\ m/s^2$

So, the the  magnitude of the acceleration is $1.41\ m/s^2$. Hence, this is the required solution.