1.

The x and y co-ordinates of a particle at any instant are x = 4t^2 + 7t and y = 5t where x, y are in metres and in seconds. The acceleration at t=5s is :

Answer»

Zero
20 `m s^(-2)`
8 `m s^(-2)`
40 `m s^(-2)`

Solution :Here a=`sqrt(a_(x)^(2)+a_(y)^(2))` But `a_(y)=0`
`( :.(d^2y)/(DT^(2))=0)` and `a=a_(x)=(d^(2)x)/(dt^(2))=8 m s^(-2)`
For any value of .t.


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