Question

A particle is moving east-wards with a velocity of 4 m/s. in 5 seconds the velocity changes to 3 m/s
northwards. Find the average acceleration in this time interval ?​

Answer 1

Answer:

1/√2 towards n-w

Explanation:

Intial velocity=v1= 5m/seast wards                                   ^||^

then after time t=10sec                                                     || towards north 5m/s

v2 =5m/s north wards                                     .⇒⇒⇒⇒⇒⇒|

                                                                towards east 5m/s

Vresultant = √v1²+v2²

                =√25+25

                 =5 √2

acceleration a=velocity/time

                     = 5√2/10 

                     =√2/2 × √2/√2

                     =1/√2 towards n-w

(direction u can understand by seeing the diagram)

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Answer 2

Answer:

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The initial velocity of the particle, $v=5i^$

After some time velocity changes to $5j^$

Change in velocity dv= $5j^ − 5i^$

Time taken t= 10 sec.

Average acceleration =

$\frac{1}{ \sqrt{2} } m {s}^{2}$

The direction is north-west making 135 degrees with east as evident from its components.

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