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

2 Answers
Sep 5, 2017

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

Sep 5, 2017

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