Question

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. `(1)/(2) ms^(-2)` towards north
B. `(1)/ (sqrt(2 ms^(-2)))` towards north - east
C. `(1)/ (sqrt (2 ms^(-2)))` towards north -west
D. ` zero`

Answer 1

Correct Answer - C
Average acceleration `= (change in velocity )/ (time interval)`
` = (Delta vec(v))/(t) `
` vecv_(1) = 5 hat(i) , vecv_(2) = 5hat(j) `
`:. Vec(a) = ( 5 hat(j) - 5 hat(i))/(10) = (hat(j))/(2)`
`:. a = sqrt( 1^(2)+(-1)^(2))/2 = sqrt(2)/2 = 1/sqrt2 ms ^(-2)`
` tan theta = (v_(2))/(v_(1)) = 5/5 = 1 :. theta = 45(@)`
Therefore the direction is North - west.