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A particle is moving eastwards with a ve...

A particle is moving eastwards with a velocity of ` 5 ms_(-1)`. In `10 seconds` the velocity changes to `5 ms^(-1)` northwards. The average acceleration in this time is

A

Zero

B

`1//sqrt(2) ms^(-2)` towards north-west

C

`1//2 ms^(-2)` towards north-west

D

`1//2 ms^(-2)` towards north.

Text Solution

Verified by Experts

The correct Answer is:
B
b. Average acceleration
`veca_(av) = (vecv(f)-vecv_(i))/t = (vecv_(f) + (-vecv_(i)))/t = vecv/t`
To find the resultant of `vecv_(f) and -vecv_(i)`, we draw the following figure.

`|vecv| = sqrt(v_(f)^(2) + v_(1)^(2)) = sqrt (5^2 + 5^2) = 5 sqrt(2) ms^(-1)`
Since, `|vecv_f| = |-vecv_(i)|`
`vecv` is directed in between `vecv_(f) and -vecv_(i)`.
Therefore, `vecv` is directed towards N-W.
`veca_(av) = (5sqrt(2))/10 = 1/(sqrt2)`.
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