A particle is moving eastwards with a velocity of ` 5 m//s`. In `10 s` the velocity changes to `5 m//s` nothwards. The average acceleration in this time is

Options

A. ` 1 /(sqrt 2) ms^(-2)` , 45 West of Noth
B. (b) ` 1/2 ms^(-2), 60^@ west of North
C. ` 2 ms^(-2) 60^@` East of sout
D. `1/(sqrt 2) ms^(-2) `, 30^@` West of south

Answer

A

Explanation

Here, initial velocity
` v_1 = 5 ms^(-1)` along East ` = 5 hat i`
` vec _1 = 5 ms^(-1)` along North `= 5 hat j`
Change in velocity,
` Deltav= vec v_2 = vec v_1 = 5 hat j - 5 hat i=- 5 hat I + 5 hat j`
Magnitude of ` Delta v= sqrt ((-5) ^2 + 5^2 ) = 5 sqrt 2 m//s`
Acceleration, ` |vec a| = (Delta v0/t = (5 sqrt 2)/(10) = 1/(sqrt 2) ms^(-2)`
` tan theta = (BD)/(OB) = 5/ 5 = 1 tan 45^@`
` theta 45^2` West of Noth`.