00:04
So the question here asks us to calculate some quantities for a particle that has a mass of 7 kgs.
00:16
And we are given that its velocity vector changes with time as 10 t square i hat plus 5 t j hat and t is in seconds and t is the variable that we work with.
00:35
So we asked to find a few quantities from this given information.
00:42
So the first question part a is solved in this case, which means that the r vector or the displacement vector, the resultant vector of this particle is, has been solved correctly to be 10 by 3 t cube ihat plus 5 by 2 t square j hat and this comes from basically integrating the v equation with time with respect time all right so this is already over the second question is that we need to describe this motion qualitatively so if you just take a look at this particle you will see that from equation a the x component of the particle is 10 by 3 t cube and the y component is 5 by 2 t square and if we eliminate t out of this we get an equation so you basically take t in terms of x and put it take t in terms of x and that put that into here when you do that you get an equation which is something like 1 .19 x square minus y cube is equal to 0 this is the relationship between the trajectory of that particle in x and y.
02:18
And if you plot this on a graph, you'll get something like this.
02:25
So qualitatively we can say that x varies non -linearly with respect to y and x changes sorry and x changes faster than y.
03:11
So the key description words are the this nonlinear relationship where x change is faster than y.
03:16
So this would be the qualitative description of this relationship.
03:28
All right.
03:29
So that explains part b.
03:32
So we come to part c which is again solved correctly in the given equation.
03:37
The acceleration is 20 t i hat plus 5 j hat and this comes by differentiating v with respect to time which is done correctly the next part also is done correctly which asks for us for the force and the force is basically mass times acceleration so it has been solved correctly by multiplying the acceleration term with the demass that is given over here you do that you get 140 t i hat plus 35 j hat so this is correct next question asks us to find the net torque now in this question we are asked for tau and not the vector so the hint here is that they're asking for the value of torque only in newton meters and we know that torque is r cross f and in this case we want the the magnitude of this value so we find our cross f over here and r cross f vector is equal to i j k where i is 10 by 3 tq j is j is 5 by 2 t squared k is 0 and and f is 140 t 35 in j and 0 and k now in this case the i and the j components will be 0 because there is multiplication with 0 if you do cross like this or cross like this both ways so only uh value you get is crossing for the k vector or the k component so this would give you the tau vector as 350 by 3 t k k 2x3 t cube and k hat so this will give us minus 233 .33 .33 k hat and so the answer would be in this case 233 .33 .33.
06:28
So the k hat is not required here...