Question

a particle has an initial velocity(6icap+8jcap)m/sec and an acceleration of (0.8icap+0.6jcap)m/secsquare.its speed after 10sec is

Answer 1

Given :-

Initial Velocity of particle = u = (6î + 8j) ms-¹

Acceleration of particle = a = (0.8î + 0.6j) ms-²

Time = t = 10 s

Now,

Using First Equation of Motion,

v = u + at

v = (6î + 8j) + (0.8î + 0.6j)10

v = (6î + 8j) + (8î + 6j)

v = (14î + 14j) ms-¹

Hence,

The velocity of particle after 10 s = v = (14î + 14j) ms-¹.

Some other Information regarding Vector :-

Important points regarding dot product :-

1. A•B = B•A

2. A•(B+C) = A•B + B•C

3. A•A = A²

4. A•B = ABCosØ

Important points regarding cross product :-

1. A × B = - B × A

2. î × î = j × j = k × k = 0

Answer 2

Answer:

Given :-

Initial Velocity of particle = u = (6î + 8j) ms-¹

Acceleration of particle = a = (0.8î + 0.6j) ms-²

Time = t = 10 s

Now,

Using First Equation of Motion,

v = u + at

v = (6î + 8j) + (0.8î + 0.6j)10

v = (6î + 8j) + (8î + 6j)

v = (14î + 14j) ms-¹

Hence,

The velocity of particle after 10 s = v = (14î + 14j) ms-¹.

Some other Information regarding Vector :-

Important points regarding dot product :-

1. A•B = B•A

2. A•(B+C) = A•B + B•C

3. A•A = A²

4. A•B = ABCosØ

Important points regarding cross product :-

1. A × B = - B × A

2. î × î = j × j = k × k = 0