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 = 26.02 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 + 1 2 at2 = 0 + 0 + (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 = 0+30 15 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: 20 + 60 + 1.67 = 81.67 sec Total distance travelled: 300 + 1800 + 25 = 2125 m Average velocity: v avg = x/ t = (2125 m)/(81.67 sec) = 26.02 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. A police car is trying to catch up to a speeder. Suppose the police car starts from rest and has an acceleration of 5 m/s 2. At that same instant, the speeder is already 25 m down the road, moving at 20 m/s away from the police car and is slowing down (i.e. deccellerating) at a rate of 1 m/s 2. How fast are the vehicles moving when the police car catches up to the speeder? (Assume everything is happening in a straight line here.) v police = 38.72 m/s and v speeder = 12.26 m/s I ll use a coordinate system with +X in the direction of motion of the two vehicles with x = 0 at the initial position of the police car. Police car : starts at the origin (x o = 0), at rest (v o = 0 ) and an acceleration of a = 5 m/s 2 so it s equation of motion will be: x P = x o +v o t+ 1 2 at2 = 0+0+ 1 2 (5)t2 or just x P = 2.5t 2. Speeder : starts at x o = 25 m with an initial velocity of v o = 20 m/s and an acceleration of a = 1 m/s 2 so it s equation of motion will be: x S = x o +v o t+ 1 2 at2 = 25+20t+ 1 2 ( 1)t2 or just x S = 25 + 20t 1 2 t2. We re interested in their speeds when they meet, at which point x P = x S so setting the two boxed equations above equal to one another: 2.5t 2 = 25 + 20t 0.5t 2. Collecting terms into the usual quadratic form, we have: 3t 2 20t 25 = 0. Applying the quadratic formula, we get two solutions: t = 1.076 s and t = +7.743 s. The police car doesn t catch up to the speeder until some positive time after it starts the chase, so t = +7.743 s must be the correct solution. Speeds : Now that we know when they meet up, we can determine their speeds at that point. v P = v o + at = 0 + 5t = 0 + (5 m/s 2 )(7.743 s) = 38.72 m/s v S = v o + at = 20 1t = (20 m/s) (1 m/s 2 )(7.743 s) = 12.26 m/s.
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 = 19.81 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 = 19.81 m/s tan θ = v oy /v ox = 17.1/10 from which θ = 59.7 o
4. A golf ball was launched with some initial speed at an angle of 40 o up from the horizontal, and it landed 100 m away. (Assume the launch point and landing points are at the same level, and the ground is horizontal and flat.) (a) How high in the air did the ball go? (I.e. what was its maximum height above the ground?) 20.98 m (b) How long was the ball in the air? 4.138 sec Since the ball launches and lands at the same height, we can use the specialized projectile motions equations here. R = v2 o sin 2θ g so rearranging: v o = Rg/ sin 2θ = (100)(9.8)/ sin 80 = 31.5455 m/s (a) How high in the air does the ball go? For this type of motion, we have h = (b) How long is the ball in the air? (vo sin θ)2 (31.5455 sin (40))2 = 2g 2 9.8 = 20.98 m. This will be twice the apogee time (the time it takes to reach the maximum height). t A = (v o sin θ)/g = (31.5455) sin (40 o )/9.8 = 2.069.. so the total time is twice this or 4.138.. sec.
5. We have two vectors in mixed units and different notations: vector A in cartesian coordinates is A = (1.1 miles, 2.2 km) vector B in polar coordinates is B = (7000 ft, 140 deg). 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 B, and the components for A so: B = B = 7000 ft 1 m = 2133 m. 3.281 ft A x = 1.1 mile 1609 m = 1770 m 1 mile A y = 2.2 km 1000 m = 2200 m 1 km Unit vector notation: (a) A = 1770 î + 2200 ĵ (meters) We were already given the components of this vector and just needed to convert their units. From earlier: A x = 1770 m and A y = 2200 m. (b) B = -1634 î + 1371 ĵ (meters) θ is measured CCW around from the +X axis, so we can directly convert polar to cartesian using: B x = B cos θ = (2133 m) cos 140 o = 1634 m B y = B sin θ = (2133 m) sin 140 o = +1371 m Polar notation: (c) A = ( 2824 m, 51.2 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 = A y /A x = 2200/1770 = 1.243.. from which θ = 51.2 o It s magnitude will be A = A 2 x + A 2 y = (1770) 2 + (2200) 2 = 2824 m. (d) B = ( 2133 m, 140 deg ) This vector was already given in polar notation, so we just needed to convert the length to meters (already done above). (e) Determine B 2A = 5995 m B 2 B = ( 1634î + 1371ĵ) 2(1770î + 2200ĵ) = 5174î 3029ĵ. The magnitude of that combination will be B 2 B = ( 5174) 2 + ( 3029) 2 = 5995.