Question

For a body starting off with an initial velocity of `vec u = 16hati + 12hatj m/s` and moving with a uniform acceleration of `vec a = -2.5hati m/s^2`, if after time 't' seconds the velocity `vec V` becomes perpendicular to it's initial velocity, then t= ?

A) 4sec
B) 4.8sec
C) 10sec
D) given condition is not possible

Answer 1

The answer is `=10s`, Option `(C)`

Explanation:

The initial velocity is `vecu= <16,12>`

The acceleration is `veca=<-2.5,0>`

We apply the equation of motion

`vecv=vecu + vecat`

Therefore,

`vecv=<16,12> + <-2.5,0>t = <16-2.5t, 12>`

As `vec v` is perpendicular to `vecu`, the dot product is `=0`

`vecu.vecv=<16,12> . <16-2.5t, 12> =16(16-2.5t)+12*12=0`

`256-40t+144=0`

`40t=256+144=400`

`t=400/40=10s`

Answer 2

`t = 10 s`, making (C) the correct solution.

Explanation:

The body moves with constant acceleration, so we can apply the vector equivalent of the "suvat" equations for motion under constant acceleration:

`{: (v=u+at, " where ", s="displacement "(m)), (s=ut+1/2at^2, , u="initial speed "(ms^-1)), (s=1/2(u+v)t, , v="final speed "(ms^-1)), (v^2=u^2+2as, , a="acceleration "(ms^-2)), (s=vt-1/2at^2, , t="time "(s)) :}`

Here we have:

`{ (s=,,m),(u=,16bb(ul hati)+12bb(ul hat j),ms^-1),(v=,bb(vecV),ms^-1),(a=,-2.5bb(ul hat i),ms^-2),(t=,t,s) :}`

Applying `bb(ul v)=bb(ul u)+bb(ul a)t` we have:

`bb(vecV) = (16bb(ul hati)+12bb(ul hat j)) + (-2.5bb(ul hat i))t`
`\ \ \ \ = (16-2.5t)bb(ul hati)+12bb(ul hat j)`

If this velocity `bb(vecV)` at time `t` is perpendicular to the initial velocity `bb(vec u)=16bb(ul hati)+12bb(ul hat j)` then we have:

`bb(vecV) * bb(vec u) = 0`

`:. (16bb(ul hati)+12bb(ul hat j)) * ((16-2.5t)bb(ul hati)+12bb(ul hat j) ) = 0`

`:. (16)(16-2.5t) + (12)(12) = 0`
`:. 16-2.5t + 9 = 0`
`:. 2.5t = 25`
`:. t = 10 \ s`