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Hi.
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Here in this given problem, initially speed of the particle in vector form that was given as v1 is equal to 4 meter per second and that is due east and its final and we should use the word velocity as it is in vector form.
00:38
So final velocity of the same particle in vector form it becomes 3 meter per second and that will be due north.
00:52
And the time taken to change this velocity that is given as 5 seconds.
01:01
So if we represent these directions, initial velocity that was due east v1.
01:11
And finally it is converted to 1.
01:19
V2 which is 3 meter per second due north like this.
01:24
So to find acceleration in vector form, it will be given as v2 minus v1 difference in the velocity per unit time.
01:44
So we need to subtract these two velocities means delta v.
01:48
That may be given as v2 minus v1 or we can say this is v2 plus negative of v1.
01:55
Here this negative of v1 will be this in vector form and now we apply vector sum triangle of a tradition this is minus v1 this is v2 so their addition will be like this angle is 90 degree so in magnitude it will simply be given by pythagore theorem means v2 square by v1 squared magnitude only so this delta v in magnitude this so this is equal to for v2 this is 3, square of 3 plus square of 4 means this is 9 plus 16 means this is 25 meter per second...