Title PSAD 5 - Kinematics.pdf ✅

PSAD 5: Kinematics Video Transcript Situation 1: A projectile is fired with an initial speed of 196 m/s at an angle 30.0 ̊ above the horizontal from the top of a cliff 98.0 m high. 1. Determine the time to reach maximum height. (Hint: vy = vy0 − gt) a. 10.5 s b. 10.0 s c. 11.5 s d. 11.0 s Solution: When analyzing projectile problems, each component of the motion is analyzed separately. Note that at the maximum height, the vertical component of velocity is 0 m/s momentarily. vy = vy0 − gt 0 = 196 sin 30° − 9.81t t = 10 s Situation 2: The tires of a truck make 60 revolutions as the car reduces its speed from 90 km/h to 50 km/h. The tires have a diameter of 0.80m. 2. What was the angular acceleration of the tires? (Hint: v0 = rω0; ωf 2 = ω0 2 + 2αθ) a. −3.58 rad/sec2 c. −5.38 rad/sec2 b. −8.53 rad/sec2 d. −5.83 rad/sec2 3. In the previous question, if the truck continues to decelerate at this rate, how much more time is required for it to stop? a. 4.1 s c. 7.6 s b. 9.7 s d. 1.4 s Solution: To convert kph to m/s, divide the values by 3.6, 90 km h × 1 m s 3.6 kph = 25 m s 50 km h × 1 m s 3.6 kph = 13.889 m s Solve for the angular velocities using v = rω. The radius of the tires is 0.8 m 2 = 0.4 m. 25 = 0.4ω1 ω1 = 62.5 rad s 13.889 = 0.4ω2 ω2 = 34.722 rad s
From the equation: ω2 2 = ω1 2 + 2αθ Note that in 1 revolution, there are 2π radians. 34.7222 = 62.5 2 + 2α(60)(2π) α = −3. 582 rad s 2 From the equation: ωf = ω0 + αt When the tire stops, that is at ωf = 0. 0 = 34.722 + (−3.582)t t = 9. 694 s Situation 3 – Refer to FIG. MECH0409 The motion of a particle starting from an initial velocity of 7 m/s is modeled in the a − t graph shown. 4. Determine the velocity (m/s) of the particle at t = 4 s. a. 14.6 b. 7.6 c. 22.2 d. 24.1 5. Determine the displacement (m) of the particle at t = 4 s. a. 10.1 b. 48.3 c. 38.1 d. 20.3 Solution: Recall that v = ∫ a dt The difference in velocities between two points in time is the area of the a-t graph between the said points in time. vt=4 = vt=0 + At=0→t=4 v4 = 7 + 1 2 (4 − 0)(3.8 − 0) v4 = 14.6 m s
Also, s = ∫ v dt = v0t + ∫ ∫ a dt For a double integral, the area moment can be used where the reference point is at t = 4. s = vt + Ax s = (7)(4) + [ 1 2 (4 − 0)(3.8 − 0)] ( 1 3 × 4) s = 38. 13 m Situation 4 – Refer to MECH99.902 To anticipate the dip and hump in the road, the driver of a car applies her brakes to produce a uniform deceleration. Her speed is 100 km/h at the bottom A of the dip and 50 km/h at the top C of the hump, which is 120 m along the road from A. If the passengers experience a total acceleration of 3 m/s2 at A and if the radius of curvature of the hump at C is 150 m, 6. Calculate the radius of curvature at A. a. 343.2 m c. 432.3 m b. 234.3 m d. 323.4 m 7. Determine the acceleration at the inflection point B. a. 2.41 m/s2 c. 1.78 m/s2 b. 1.29 m/s2 d. 2.73 m/s2 8. Determine the total acceleration at C. a. 2.41 m/s2 c. 1.78 m/s2 b. 1.29 m/s2 d. 2.73 m/s2 Solution: Convert the speeds to m/s 100 km h × 1 m s 3.6 kph = 27.778 m s 50 km h × 1 m/s 3.6 kph = 13.889 m s Calculate the tangential acceleration of the car, v2 2 = v1 2 + 2as, at = 13.8892 − 27.7782 2(120) = −2.411 m s 2
From Pythagorean theorem, the relationship between the tangential and the normal acceleration is a = √an 2 + at 2 3 = √an 2 + (−2.411) 2 an = 1.785 m s 2 From the formula for the normal acceleration, an = v 2 r 1.785 = 27.7782 rA rA = 432. 30 m Since curvature changes at the inflection point, then the radius of curvature approaches infinity. This means that an = 0 at point B. Therefore, the acceleration at point B is just the tangential acceleration, at = −2. 411 m s 2 . Note that the tangential acceleration is constant all throughout. Compute for the normal acceleration at point C. an = v 2 r an = 13.8892 150 = 1.286 m s 2 Therefore, the total acceleration is a = √an 2 + at 2 a = √1.2862 + (−2.411) 2 a = 2. 733 m s 2 Situation 5: The position of a particle which moves along a straight line is defined by the relation x = t 3 − 6t 2 − 15t + 40 where x is in meters and t is in seconds. 9. Determine the time at which the velocity will be zero. a. 5.0 s c. 1.8 s b. 7.3 s d. 3.1 s 10. Determine the position of the particle at that time. a. −60 m c. −34 m b. 60 m d. 34 m 11. Determine the velocity of the particle when its acceleration is zero. a. −27 m/s c. −6 m/s b. 0 m/s d. 2 m/s