Velocity and Acceleration

Level pending

A particle moves in a 3 3 -dimensional space so that its acceleration vector at any time t > 0 t>0 is a ( t ) = − 4 i − 6 t j + 12 t k \boldsymbol{a}(t) = -4\ \boldsymbol{i} -6t\ \boldsymbol{j}+12t\ \boldsymbol{k} , initial velocity is v ( 0 ) = 0 \boldsymbol{v}(0) = 0 , and initial point is r ( 0 ) = j − 2 k \boldsymbol{r}(0) = \boldsymbol{j}-2\ \boldsymbol{k} . The position vector at t = 4 t = 4 is represented by r ( 4 ) = a i + b j + c k \boldsymbol{r}(4) = a\ \boldsymbol{i} + b\ \boldsymbol{j} + c\ \boldsymbol{k} . What is the value of a + b + c a+b+c ?

31 31 21 21 41 41 11 11

Ajay Dutta by Ajay Dutta , 24, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Ashtamoorthy Ts
Mar 7, 2014

v(t) = integral a(t) dt ; v(t=0) = 0 fixes the constant of integration to zero. r(t) = integral v(t) dt; r(0) = j - 2k fixes the constant of integration to j-2k r(4) = ai+bj+ck; a comparison gives a+b+c. note: r, v, a are position, velocity, and acceleration vectors.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...