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An object has a velocity, v = (2hati + 4...

An object has a velocity, `v = (2hati + 4hatj) ms^(-1)` at time `t = 0s`. It undergoes a constant acceleration `a = (hati - 3hatj)ms^(-2)` for 4s. Then
(i) Find the coordinates of the object if it is at origin at `t = 0`
(ii) Find the magnitude of its velocity at the end of 4s.

A

`vec a= 2hatj+hatk`

B

`veca = 7hati-6hatj-hatk`

C

`veca = 7hati+hatj-hatk`

D

`veca = 3hati-4hatj+2hatk`

Text Solution

Verified by Experts

The correct Answer is:
B
An object ……………
speed starts decreasing
of `veca. vecv lt 0`
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