To find the velocity of the particle at t=5s, we can use the kinematic equation:
v = u + at
where: v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time interval.
Given: u° = -2i + 4j m/s (initial velocity) a° = 3 m/s^2 (magnitude of acceleration) angle between a° and x-axis = 130°
First, we need to express the acceleration vector in terms of unit vectors. The angle between a° and the x-axis is 130°, so the angle between the acceleration vector a and the positive x-axis is 130° - 180° = -50°.
Using the magnitude of the acceleration and the angle, we can calculate the components of the acceleration vector:
ax = a° * cos(angle) ay = a° * sin(angle)
Substituting the values, we have:
ax = 3 * cos(-50°) ay = 3 * sin(-50°)
Now, let's calculate the components of the final velocity vector at t=5s:
vx = u°x + ax * t vy = u°y + ay * t
Given that t = 5s, we can substitute the values:
vx = -2 + ax * 5 vy = 4 + ay * 5
Now, we need to find the unit vector notation for the velocity vector. The unit vector notation represents a vector in terms of its direction. The unit vector is found by dividing the vector by its magnitude.
The magnitude of the velocity vector is given by:
|v| = √(vx^2 + vy^2)
Substituting the values, we have:
|v| = √((-2 + ax * 5)^2 + (4 + ay * 5)^2)
Finally, we can express the velocity vector v° at t=5s in unit vector notation:
v° = (vx/|v|)i + (vy/|v|)j
Calculating the above expressions will give us the final velocity v° at t=5s in both unit vector notation and magnitude.