Use significant figures to calculate the sum of the following numbers; \( 12,35+7,8+19,071+5,009 \) A car starts from rest and reaches a speed of \( 90 \mathrm{~km} / \mathrm{h} \) in 10 seconds: a) Convert this speed to SI units. b) Calculate its average acceleration. c) Calculate it final displacement at \( \mathrm{t}=5 \mathrm{~s} \).

Below is a step‐by‐step solution. ───────────────────────────── 1. Sum Using Significant Figures We are given the numbers:   12.35  (4 significant figures, two decimal places)   7.8   (2 significant figures, one decimal place)   19.071 (5 significant figures, three decimal places)   5.009  (4 significant figures, three decimal places) When adding numbers the answer is reported to the least number of decimal places among the addends. Here the least number of decimal places is one (from 7.8). Step 1.1: Compute the sum exactly:   12.35 + 7.8 = 20.15   20.15 + 19.071 = 39.221   39.221 + 5.009 = 44.230 Step 1.2: Round the result to one decimal place:   44.230 → 44.2 Thus, the sum is 44.2. ───────────────────────────── 2. Car Motion Problem A car starts from rest and reaches 90 km/h in 10 s. ───────────────────────────── (a) Convert 90 km/h to SI Units Convert km/h to m/s by using the conversion factor 1 km/h = 1000 m/3600 s = 1/3.6 m/s:   90 km/h = 90 ÷ 3.6 = 25.0 m/s (We write 25.0 m/s to indicate that the conversion is carried out with appropriate precision.) ───────────────────────────── (b) Calculate the average acceleration The car starts from rest (v₀ = 0) and reaches v = 25.0 m/s in t = 10 s. The average acceleration is given by:   a = (v - v₀) / t = (25.0 m/s - 0) / 10 s = 2.50 m/s² Rounded appropriately (using 2 or 3 significant figures as in the given speed), we take the average acceleration as 2.5 m/s². ───────────────────────────── (c) Calculate its displacement at t = 5 s For uniformly accelerated motion from rest, the displacement is given by:   x = v₀t + ½ a t² Since v₀ = 0, then:   x = ½ a t² = ½ × 2.5 m/s² × (5 s)²   x = 1.25 m/s² × 25 s² = 31.25 m Now, because the acceleration (2.5 m/s²) is known to 2 significant figures and the time is given as 5 s (assumed exact or at least sufficiently precise), the result should be rounded to 2 significant figures. Thus, 31.25 m rounds to 31 m. ───────────────────────────── Final Answers 1. Sum of the numbers: 44.2 2(a). 90 km/h = 25.0 m/s 2(b). Average acceleration = 2.5 m/s² 2(c). Displacement at t = 5 s = 31 m