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
A) 4sec
B) 4.8sec
C) 10sec
D) given condition is not possible
2 Answers
The answer is
Explanation:
The initial velocity is
The acceleration is
We apply the equation of motion
Therefore,
As
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(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) * 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