A particle travels to the right at a constant rate of 6.8 m/s. It suddenly is given a vertical acceleration of 1.8 m/s2 for 5 s. What is its direction of travel after the acceleration with respect to the horizontal? Answer between −180◦ and +180◦
and
What is the speed at this time? Answer in units of m/s. Answer in units of ◦
To determine the direction of travel after the acceleration, we need to consider both the horizontal and vertical components of the particle's motion.
Initially, the particle is traveling to the right, which means its horizontal component is positive. After the vertical acceleration is applied, we need to find the new direction by calculating the angle between the new velocity vector and the horizontal axis.
To find this angle, we can use trigonometry. The tangent of the angle is given by the ratio of the vertical component to the horizontal component:
tan(θ) = vertical component / horizontal component
The vertical component can be determined using the equation for uniformly accelerated motion:
Vertical component = initial velocity * time + 0.5 * acceleration * time^2
Initially, the vertical component is zero since the particle is only moving horizontally. The acceleration is 1.8 m/s^2, and the time is 5 s. Thus, the vertical component is:
Vertical component = 0 + 0.5 * 1.8 * (5^2) = 0 + 0.5 * 1.8 * 25 = 22.5 m/s
The horizontal component remains constant at 6.8 m/s. Now we can find the angle θ by taking the inverse tangent:
θ = arctan(Vertical component / horizontal component)
θ = arctan(22.5 / 6.8)
Using a calculator, we find that θ is approximately 73.32°.
Since the particle initially travels to the right, its direction of travel after the acceleration is 73.32° to the right of the horizontal axis. The answer should be in the range of -180° to +180°, so the final answer is +73.32°.
To find the speed at this time, we can use the Pythagorean theorem. The speed is the magnitude of the particles' velocity vector, which is the square root of the sum of the squares of the horizontal and vertical components:
Speed = sqrt(horizontal component^2 + vertical component^2)
Speed = sqrt(6.8^2 + 22.5^2)
Using a calculator, we find that the speed is approximately 23.55 m/s.
Therefore, the direction of travel after the acceleration is +73.32° with respect to the horizontal axis, and the speed at this time is approximately 23.55 m/s.