A particle's constant acceleration is south at 2.1 m/s2. At t = 0, its velocity is 45.2 m/s east. What is its velocity at t = 7.7 s? What's the magnitude and direction?
Vf=Vi+at
Vf=45.2E+ 2.1*7.7 S
I got 61.37 but I'm getting it wrong
To determine the particle's velocity at t = 7.7 s, we need to 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.
Given:
Initial velocity (u) = 45.2 m/s (east)
Acceleration (a) = 2.1 m/s^2 (south)
Time (t) = 7.7 s
Since velocity is a vector quantity, we need to consider both magnitude and direction. The magnitude is the numerical value of the velocity, while the direction indicates its orientation.
Step 1: Determine the final velocity in the east-west direction (x-component).
The acceleration is in the south direction, so it does not affect the particle's velocity in the east-west direction. Hence, the x-component of the velocity remains unchanged.
v_x = u_x = 45.2 m/s
Step 2: Determine the final velocity in the north-south direction (y-component).
To determine the y-component of the final velocity (v_y), we need to consider the southward acceleration acting for 7.7 seconds.
Using the formula: v_y = u_y + a_yt
Since there is no initial velocity in the north-south direction (u_y = 0), the equation simplifies to:
v_y = a_yt
v_y = (2.1 m/s^2) * (7.7 s) = 16.17 m/s (south)
Step 3: Combine the x-component and y-component of the final velocity to obtain the magnitude and direction.
Using the Pythagorean theorem, we can find the magnitude (v) of the final velocity.
v = sqrt(v_x^2 + v_y^2)
= sqrt((45.2 m/s)^2 + (16.17 m/s)^2)
= sqrt(2046.04 m^2/s^2 + 261.6489 m^2/s^2)
= sqrt(2307.6889 m^2/s^2)
= 48.04 m/s (approximately)
The magnitude of the final velocity is approximately 48.04 m/s.
To determine the direction, we need to find the angle between the final velocity vector and the east direction using trigonometry.
θ = atan(v_y / v_x)
= atan(16.17 m/s / 45.2 m/s)
= atan(0.3578)
≈ 19.1° (south of east)
Therefore, the final velocity at t = 7.7 s is approximately 48.04 m/s with a direction of 19.1° south of east.