A particle moving at a velocity of 8 m/s in the positive x direction is given an acceleration of 7.4 m/s2 in the positive y direction for 1.7 s.
What is the final speed of the particle? Answer in units of m/s
Vy = 7.4*1.7 = 12.58 m/s
So, the final speed is
√(8^2+12.58^2) = 14.91 m/s
To find the final speed of the particle, we need to calculate the magnitude of the velocity vector after the given acceleration is applied.
First, let's analyze the magnitude of the initial velocity vector. We are told that the particle is moving at a velocity of 8 m/s in the positive x direction. The magnitude of this velocity vector is simply the absolute value of the velocity, which is 8 m/s.
Now, let's analyze the magnitude of the acceleration vector. We are given that the particle experiences an acceleration of 7.4 m/s^2 in the positive y direction. The magnitude of this acceleration vector is also 7.4 m/s^2.
Since the initial velocity and the acceleration are in different directions (x and y directions, respectively), we need to combine them to determine the resulting velocity vector.
To do this, we can use vector addition. The magnitude of the resultant velocity vector can be found using the Pythagorean theorem:
resultant magnitude = √(initial velocity magnitude^2 + acceleration magnitude^2)
= √((8 m/s)^2 + (7.4 m/s^2)^2)
Calculating this results in:
resultant magnitude = √(64 m^2/s^2 + 54.76 m^2/s^4)
= √118.76 m^2/s^2
≈ 10.9 m/s
Therefore, the final speed of the particle is approximately 10.9 m/s.