Solve uniform-acceleration and free-fall problems with the kinematic equations, including overtaking, reaction time and multi-part problems

Learning goal: Use the definitions of average velocity and acceleration, and the kinematic equations they lead to, to solve problems about a body accelerating uniformly in one dimension, including free fall with negligible air resistance; use the answer from one part of a problem in the next; solve simultaneous equations for when an accelerating object overtakes one at constant velocity; check units, significant figures and limiting cases; and determine g, reaction time or jump speed from measured data.

Where the equations come from. For any motion, average velocity is displacement over time, v_avg = Δx/Δt, and average acceleration is change in velocity over time, a = Δv/Δt. When acceleration is constant, the velocity graph is a straight line, so the average velocity is the mean of the initial and final velocities, v_avg = (v_i + v_f)/2. That shortcut fails if the acceleration changes, because the graph is then no longer straight. From these three relations come the equations used in this course, written with v_i for initial velocity, v_f for final velocity, a for acceleration, t for elapsed time and Δx for displacement:

v_f = v_i + at; Δx = ½(v_i + v_f)t; Δx = v_i t + ½at²; v_f² = v_i² + 2aΔx.

Each equation leaves out one quantity. The first has no Δx, the second no a, the third no v_f and the fourth no t. Choosing an equation means listing what you know, what you want, and which quantity you can ignore. Every quantity is a signed number along one chosen positive direction; velocity and acceleration can be negative, and Δx is displacement, not distance.

Free fall. Near Earth's surface, with air resistance negligible, every freely falling object has the same downward acceleration, about 9.8 m/s² (a more precise value is 9.81 m/s²; this course uses 9.8). "Freely falling" includes objects on the way up: a ball thrown upward is in free fall from the moment it leaves the hand, its velocity decreasing by 9.8 m/s each second, passing through zero at the top, and then growing in the negative direction. Take up as positive and a = −9.8 m/s². At the highest point the velocity is zero but the acceleration is still −9.8 m/s²: gravity keeps pulling on the ball there, so its velocity is zero only for an instant and is already changing toward negative values. Zero velocity at an instant does not mean zero acceleration.

A fully worked model. A ball is thrown straight up at 14.7 m/s from a hand at ground level. (a) How long does it take to reach the top? At the top v_f = 0, so 0 = 14.7 + (−9.8)t and t = 1.5 s. (b) How high does it go? Using v_f² = v_i² + 2aΔx: 0 = 14.7² + 2(−9.8)Δx, so Δx = 216.09 ÷ 19.6 = 11.0 m. Check with the other equation: Δx = ½(14.7 + 0)(1.5) = 11.0 m. (c) When does it return to the hand? Δx = 0 = 14.7t − 4.9t², so t(14.7 − 4.9t) = 0 and t = 3.0 s, twice the time to the top, as symmetry requires. (d) Its velocity then: v_f = 14.7 − 9.8 × 3.0 = −14.7 m/s, the same speed downward. Part (c) used the answer of part (a) as a check; that is the habit of using each part's result in the next.

Units, significant figures and limiting cases. Carry units through every line: 14.7² gives m²/s², and dividing by 19.6 m/s² leaves meters. Report answers with the precision of the data: 14.7 m/s has three significant figures, so 11.0 m, not 11.025 m. Then test the answer against a limit. If g were larger, the ball should go less high; the formula v_i²/(2g) does that. If v_i were zero, the height should be zero; it is. A formula that misbehaves in an easy limit is wrong even if the number looks plausible.

Overtaking problems. A speeder passes a parked police car at a constant 24 m/s. At that instant the police car starts from rest with an acceleration of 3.0 m/s². Where and when is the speeder caught? Write both positions from the same origin and clock: x_speeder = 24t and x_police = ½(3.0)t² = 1.5t². Setting them equal, 24t = 1.5t², and dividing by t (t = 0 is only the starting moment) gives t = 16 s, x = 384 m. Now the constraint: at 16 s the police car would be moving at 3.0 × 16 = 48 m/s, which is above its maximum of 36 m/s. The first answer is therefore invalid and the problem must be solved in stages. The police car reaches 36 m/s at t = 12 s, having covered 1.5 × 144 = 216 m; the speeder is then at 24 × 12 = 288 m, 72 m ahead. From then on the gap closes at 36 − 24 = 12 m/s, taking 72 ÷ 12 = 6.0 s more. The catch is at t = 18 s and x = 24 × 18 = 432 m; check: 216 + 36 × 6.0 = 432 m. A reaction time before the police car moves would be handled the same way, as another stage. Problems in this course are set so that the two equations can be solved by factoring, by working in stages, or by reading where the two position graphs cross; problems that need the quadratic formula are beyond what this course asks. A graph of both positions is always a useful check.

Measured data. Reaction time: a partner holds a ruler vertically with your fingers open at the 0 cm mark; when it drops, you catch it, and it falls d before you close. From Δx = ½gt², t = √(2d/g). A catch at 18 cm gives t = √(2 × 0.18 ÷ 9.8) = 0.19 s. Jump: if a jumper's feet are off the ground for 0.60 s, the rise takes 0.30 s; launch speed is v = gt = 9.8 × 0.30 = 2.9 m/s and height is ½ × 9.8 × 0.30² = 0.44 m. The ratio of rise time to fall time is 1:1 only if the center of mass is at the same height at take-off and landing, which needs the same body posture at both moments.

Investigation: determine g. Materials: a steel ball or golf ball, a tape measure, a smartphone video at 60 frames per second or a ticker timer, and a floor mat. Steps: release the ball from rest at heights of 0.50, 1.00, 1.50 and 2.00 m and time each fall from the frame of release to the frame of impact; repeat three times per height. Record height and mean time. Analyze: since Δx = ½gt², plotting height against t² gives a straight line through the origin whose slope is g/2. A result that supports the claim: a slope near 4.9 m/s², so g near 9.8 m/s². For example, a measured time of 0.66 s for 2.00 m gives g = 2 × 2.00 ÷ 0.66² = 9.2 m/s², a 6 percent shortfall. Sources of error: release and impact frames are each uncertain by one frame (0.017 s), which matters most for the shortest drop; a slight push at release; air drag on a light ball. Improvements: use a denser ball, more heights, a higher frame rate or a photogate, and take the slope of the whole plot instead of one point. Safety: drop onto a mat, not toes. What the model cannot show: the graph assumes air resistance is negligible, which holds for a dense ball over 2 m but would not for a table-tennis ball.

Worked example 1 — a car in two stages. A car starts from rest and accelerates at 2.5 m/s² for 8.0 s, then brakes at −5.0 m/s² to a stop. Find the total distance. Stage 1: v_f = 0 + 2.5 × 8.0 = 20 m/s and Δx = ½ × 2.5 × 8.0² = 80 m. Stage 2 uses the 20 m/s from stage 1: v_f² = v_i² + 2aΔx gives 0 = 400 + 2(−5.0)Δx, so Δx = 40 m, and the braking time is (0 − 20) ÷ (−5.0) = 4.0 s. Total distance 120 m in 12 s. Check: the average speed over the braking is 10 m/s for 4.0 s, 40 m.

Worked example 2 — a stone dropped down a well. A stone dropped from rest takes 2.4 s to hit the water. How deep is the well, and how fast is the stone moving at impact? Take down as positive: Δx = 0 + ½ × 9.8 × 2.4² = 28 m (28.2 before rounding to the two significant figures of 2.4 s). Impact speed v_f = 9.8 × 2.4 = 24 m/s (23.5). Check the limit: a 1 s drop would be 4.9 m, and 2.4 s is more than twice as long, so more than four times as deep; 28 m is consistent.

Worked example 3 — an accelerating cart overtakes a constant-velocity buggy. A buggy moves along a track at 0.50 m/s. A cart released from rest 1.5 s later at the same start line accelerates at 0.50 m/s². When and where does the cart catch the buggy? Use the clock of the cart's release: x_buggy = 0.75 + 0.50t (it had a 0.75 m head start) and x_cart = ½ × 0.50 × t² = 0.25t². Setting them equal gives 0.25t² − 0.50t − 0.75 = 0; multiplying by 4, t² − 2t − 3 = 0, which factors as (t − 3)(t + 1) = 0. The root t = −1 s is rejected, because the cart has not been released before t = 0, so t = 3.0 s. Position: x = 0.75 + 0.50 × 3.0 = 2.25 m (check: 0.25 × 3.0² = 2.25 m). A graph of both positions confirms it: at t = 2.0 s the buggy (1.75 m) is ahead of the cart (1.0 m), and at t = 4.0 s the cart (4.0 m) is ahead of the buggy (2.75 m).

Worked example 4 — velocity of a jump from time in the air. A student's feet leave the ground and return 0.50 s later, landing in the same posture as at take-off, so the body's center of mass is at the same height at both moments. Time to the top is 0.25 s. Take-off speed is v = 9.8 × 0.25 = 2.5 m/s (2.45). Height is ½ × 9.8 × 0.25² = 0.31 m (0.306). Check: using v_f² = v_i² + 2aΔx, 0 = 2.45² − 19.6Δx gives Δx = 0.306 m, agreeing.

Representation task: an equation map. Make a table with five columns headed v_i, v_f, a, t and Δx. For each of the four kinematic equations, check off the four quantities it contains and cross out the one it omits. Below the table, solve the model problem from the lesson (a ball thrown up at 14.7 m/s) a second time using different equations from the ones used in the text for parts (a) and (b), and show that the results agree. Then draw the velocity vs. time graph for the whole flight and label the top of the flight, the two areas and their signs.

Independent performance: predict an overtake, then test it. With a constant-velocity buggy and a cart on a low ramp, measure the buggy's velocity from a position vs. time graph and the cart's acceleration from at least four position-time points (use the tangent or the fit x = ½at²). Predict, by writing and solving the two position equations, when and where the cart will pass the buggy if the cart is released 1.5 s after the buggy crosses the start line. Run the test twice and record where the pass happened. Success criteria: both motion equations are written with the same origin and clock; the prediction states a time and a position with units and two significant figures; the measured result is compared with the prediction as a percentage difference; and at least two physical reasons for the discrepancy are given (for example, the buggy slowing as its battery drains, friction on the ramp, or timing uncertainty).