PH Fall - Section 04 - Version A DRAFT

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1 1. A truck (traveling in a straight line), starts from rest and accelerates to 30 m/s in 20 seconds. It cruises along at that constant speed for one minute, then brakes, coming to a stop in 25 m. Determine the vehicle s average velocity (in meters/sec) over this entire event (from start to stop). (a) v avg = m/s Coordinate system: the truck is moving in a straight line, so let s define that direction to be the +X axis. We have three different acceleration segments here, so need to look at each separately, computing the time and distance travelled in each segment so that we can determine the total distance and total time for this scenario, which we need to determine the overall average velocity: v avg = x/ t. Acceleration segment : object starts at rest and reaches 30 m/s in 20 sec. v = v o + at so 30 = 0 + (a)(20) or a = 1.5 m/s 2. Distance travelled: x = x o + v o t at2 = (0.5)(1.5)(20) 2 = 300 m. Shorter option: x = v avg t. Over this segment, we have an average velocity of v avg = m/s, so x = (15 m/s)(20 s) = 300 m. 2 = Cruising segment : object is moving at a constant velocity of 30 m/s for 1 minute (i.e. 60 seconds) so it travels a distance of (30 m/s)(60 s) = 1800 m. Braking segment : the object drops from 30 m/s to 0 m/s over a distance of 25 m. The average velocity over just this segment is v avg = 30+0 = 15 m/s so x = v 2 avg t for this segment implies that (25 m) = (15 m/s) t from which t = 1.67 sec. Overall average velocity Accumulating the information from the three segments above: Total time: = sec Total distance travelled: = 2125 m Average velocity: v avg = x/ t = (2125 m)/(81.67 sec) = m/s. (Note: the long segment in the middle where the velocity was a constant 30 m/s represented most of the time here, so the average velocity should be just a little below that.)

2 2. In the movie Unstoppable, a truck is trying to catch up with a runaway train. Suppose the truck starts from rest and has an acceleration of 4 m/s 2. When the truck takes off, the train is already 100 m down the track, moving at 15 m/s away from the truck and the train is deccelerating (i.e. slowing down) at a rate of 1 m/s 2. How fast are the vehicles moving when the truck catches up to the train? (Assume everything is happening in a straight line here.) v truck = 40.0 m/s and v train = 5.00 m/s If we start t = 0 when the truck starts accelerating after the train, both vehicles have constant acceleration, so we don t need to do any offsetting of the time axis here. Truck position: x truck = x o + v o t at2 = (4)t2 = 2t 2. Train position: x train = x o + v o t at2 = t ( 1)t2 = t 0.5t 2. When the truck catches up to the train, they ll have the same x coordinate, so setting these two equations equal to one another, we have: 2t 2 = t 0.5t 2 or rearranging: 2.5t 2 15t 100 = 0. This quadratic equation has two solutions, t = 4.00 s and t = s. It obviously takes some positive amount of time for the truck to catch up to the train, so t = s must be the actual solution. Velocities v truck = v o + at = 0 + (4 m/s 2 )t = 0 + (4 m/s 2 )(10 s) = 40.0 m/s v train = v o + at = (15 m/s) + ( 1 m/s 2 )t = (15 m/s) + ( 1 m/s 2 )(10 s) = 5.00 m/s.

3 3. We kick a soccer ball from the edge of the roof of Hilbun and observe that it lands on the ground exactly 4 seconds later, after travelling a horizontal distance of 40 meters. If the roof is exactly 10 meters above the ground level, what must have been the initial velocity of the soccer ball (in terms of it s speed and angle relative to the horizontal)? v o = m/s and θ = 59.7 deg ADD FIGURE The ball launches and lands at different heights, so we can t use the specialized projectile motion equations: this one we ll just have to brute force starting with the generic equations of motion. Define a coordinate system with the origin where the ball was launched, with +Y vertically upward and +X horizontal and to the right. X direction x = x o + v ox t + 1a 2 xt 2. By our choice of origin, x o = 0 and we have no acceleration in the X direction so this reduces to just: x = v ox t. We know that at t = 4 the ball lands at x = 40 m so: (40) = (v ox )(4) or v ox = 10 m/s. Y direction y = y o + v oy t + 1a 2 yt 2. By our choice of origin, y o = 0 and we have the acceleration due to gravity, which is downward so a y = g = 9.8 m/s 2 so this equation reduces to: y = v oy t 4.9t 2. We know that at t = 4 the ball lands at y = 10 so: ( 10) = (v oy )(4) (4.9)(4 2 ) from which v oy = 17.1 m/s. Initial Speed and Direction of ball v o = v 2 ox + v 2 oy = (10) 2 + (17.1) 2 = m/s tan θ = v oy /v ox = 17.1/10 from which θ = 59.7 o

4 4. We want to design a stunt where a motorcycle jumps over a series of cars. The motorcycle will be launched at some unknown speed at an angle of 30 o up from the horizontal and we need it to land exactly 60 m away. (Assume the launch point and landing points are at the same level, and the ground is horizontal and flat.) (a) How long will the motorcycle be in the air? 2.66 sec (b) How high in the air will the motorcycle go? (I.e. what will be its maximum height above the ground?) 8.66 m Since we re launching and landing at the same elevation, we can use the specialized projectile motion equations. In particular: sin 2θ R = v2 o so v g o = (Rg)/ sin 2θ with R = 60 m, g = 9.8 m/s 2 and θ = 30 o : v o = (60 9.8)/ sin 60 = m/s. (a) How long is the motorcycle in the air? vo sin θ For this type of trajectory, the total time will be 2t a where t a = so the total time from launch to landing will be twice that or s (b) Maximum height above ground. g = The maximum height will be h = 1 (v 2g o sin θ) 2 = 1 (26.06 sin )2 = 8.66 m (26.06) sin = s

5 5. We have two vectors in mixed units and different notations: vector A in polar coordinates is A = (2000 ft, 150 deg) vector B in cartesian coordinates is B = (1.5 km, 0.7 mile) Write each vector in unit vector and polar notations, using units of meters for all lengths, and be sure to SHOW all your calculations and units conversions: UNITS CONVERSIONS : let s do these first. We were given the magnitude (length) of A, and the components for B so: A = A = 2000 ft 1 m = m ft B x = 1.5 km 1000 m = 1500 m 1 km B y = 0.7 mile 1609 m = 1126 m 1 mile Unit vector notation: (a) A = î ĵ (meters) θ is measured CCW around from the +X axis, so we can directly convert polar to cartesian using: A x = A cos θ = (609.6 m) cos 150 o = m A y = A sin θ = (609.6 m) sin 150 o = m (b) B = î ĵ (meters) We were already given the components of this vector and just needed to convert their units. From earlier: B x = 1500 m and B y = 1126 m. Polar notation: (c) A = ( m, 150 deg ) This vector was already given in polar notation, so we just needed to convert the length to meters. (d) B = ( 1876 m, deg ) The X and Y components of this vector were both positive, so it s in the first quadrant. We can unambiguously determine the angle from tan θ = B y /B x = 1126/1500 = from which θ = o It s magnitude will be B = Bx 2 + By 2 = (1500) 2 + (1126) 2 = 1876 m. (e) Determine A 2B = 4030 m A 2B = ( 527.9î ĵ) 2(1500î ĵ) = 3528î 1947ĵ The magnitude of that combination will be A 2B = ( 3528) 2 + ( 1947) 2 = 4030.