Question

1. At t=0s, a particle moving in the x-y plane with constant acceleration has a velocity ofv; = (3î-2)) m/s, and is at the origin. At t=3s, the particle's velocity is f = (91+7j) m/s. Find (a)the acceleration of the particle (b) Its coordinates at t=3s​

Answer 1

We can use the kinematic equations of motion to solve this problem.

Let the acceleration of the particle be a = axî + ayj.

(a) Using the equation of motion v = u + at, where u is the initial velocity:

f = v = u + at

Substituting the given values, we get:

(91+7j) = (3î-2j) + a(3î + 3j)

Equating the real and imaginary parts, we get:

91 = 3a + 3a (coefficients of î are equated)

7 = -2a + 3a (coefficients of j are equated)

Solving these equations simultaneously, we get:

a = î(23/6) + j(1/2)

So the acceleration of the particle is a = (23/6)î + (1/2)j.

(b) Using the equation of motion s = ut + (1/2)at^2, where s is the displacement and u is the initial velocity:

At t = 3s, the displacement of the particle is:

s = ut + (1/2)at^2

Substituting the given values, we get:

s = (3î-2j)(3) + (1/2)(23/6)î(3)^2 + (1/2)(1/2)j(3)^2

Simplifying, we get:

s = 9î + (17/2)j

So the coordinates of the particle at t=3s are (9, 17/2).

Answer 2

the particle is at coordinates (18,15/2)

Step-by-step explanation:

To find the acceleration of the particle, we can use the formula for velocity: v = v0 + at, where v0 is the initial velocity, a is the acceleration, and t is the time. Since we know the initial and final velocities, as well as the time interval, we can solve for the acceleration:

a = (v - v0)/t = [(9i + 7j) - (3i - 2j)]/3 = (6i + 9j)/3 = 2i + 3j

So the acceleration of the particle is a = 2i + 3j m/s².

To find the coordinates of the particle at t=3s, we can use the formula for position: r = r0 + v0t + 1/2at², where r0 is the initial position. Since the particle starts at the origin, r0 = 0. Plugging in the values we have:

r = 0 + (3i - 2j)(3) + 1/2(2i + 3j)(3)² = 9i - 6j + 9i + 27/2 j = 18i + 15/2 j