pascals programming lan
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1. The document contains a student's solutions to physics problems involving motion. It includes calculations of speed, velocity, and average speed for scenarios like a tortoise and hare race, motion of a particle, speed of a slow crab, and average speed for a car trip with varying speeds. 2. The student provides detailed calculations and step-by-step working to solve for values requested in each problem, such as time taken, velocity, speed, or average speed. 3. The problems cover concepts like constant acceleration, varying speeds, and calculating average rates over a full distance or time period.
pascals programming lan
- 1. NAME: ALAKA PRECIOUS CHINONSO COURSE: ENG102 DEPARTMENT: ENGINEERING (CIVIL) LEMASS QUESTION 1. A tortoise and a hare are in a road race to defined the honour of their breed. The tortoise craws the entire 1000m at a speed of 0.2m/s. the rabbit runs the first 200meters at 2m/s, stops to take a nap for 1.3hrs, and awakens to finish the last 800meters with an average speed of 3m/s. who wins the race and by how much time? Solution Tortoise distance=1000m, speed=0.2m/s Rabbit distance=200meters, speed=2m/s Time for taking a nap=1.3*3600=4,680secs After the nap distance covered=800meters,average speed of 3m/s. Tortoise= sav =distance/time. 0.2=1000/t 0.2t=1000 T=1000/2 Therefore t=5,000secs Rabbit=sav=distance/time 2=200/t
- 2. 2t=200 Therefore t=100seconds After nap time=1.3hrs=1.3*3600=4680 =4680+100=4780 sav of rabbit after nap, sav=distance/time 3=800/t 3t=800 T=800/3=266.67 T=266.67s Total time taken by rabbit=266.67+4780 =5046.67secs Total time= (5046.67-5000) =46.67seconds. Therefore the rabbit won the race at 46.67sec ahead of rabbit. 2. If a particle’s position is given by x=6-12+14t2 (Where t is in secs and x is in meters) 1. What is the velocity at 1s 2. What is the speed at 1s 3. is there ever an instant when the velocity is o? if so give the time. Solution X=6-12t+14t2 V=?
- 3. S=? X=6-12*1+14(1) X=6-12+14 X=-6+14 X=8 At t=1s, speed =av/t=8/1=8 Speed=8 X=6-12*1+14 X=6-12+14 X=-6+14 X=8 At t=1s velocity=8m/s There is an instant when v=o Therefore time=1s 3. The slowest animal ever discovered is a crab found in the red sea that travels an average speed of 5.7km/year. How long will it take this crab to travel 1meter? Solution sav =5.7km/year Time to take the crab=? Distance=1m 5.7km=5.7*1000=5700m 1yr=5700m sav=distance/time
- 4. 5700=1/t 5700t=1 T=1/5700 T=0.000175seconds. 4. A car travels up a hill at a constant speed of 37km/h and returns down the hill at a constant of 66km/h. calc. the average speed for the whole trip. Solution V1=37km/hr V2=66km/h The first time,t1=d/2/v1 D /2 *1/v1 d/2*1/37=d/74 t2=d/2/v2 D /2 *1/v2 D /2 * 1/66=d/132 T2=d/132 Sum t1+t2 d/37+d/132 L.c.m of 37 and 132=4884 37d+66d/4884 103d/4884 Total time=0.0211d
- 5. sav =d/0.0211d sav=2/0.211*2 sav=2/0.0422 sav=43.4km/hr 5. We drive a distance of 1km at 16km/h. then we drive an additional distance of 1km at 32km/h. what is our average speed Solution Distance of 1km at 16km/hour Additional distance of 1km at 32km/hour t1=d/2/t t1=16/2*/t t1=8/t t2=d/2/t t2=32/2*1/t t2=16/t sum 8/t + 16/t =16 + 8/t=24/t sav =distance/time sav=24/t sav=24/2 sav=12km/hour 12*1000/3600=1200/3600
- 6. 3.333m/s 6. A car is initially travelling due north at 23m/s A. find the velocity of the car after 4s if it’s acceleration is 2m/s2 due north. B. find the velocity of the car after 4s if it’s acceleration is instead 2m/s2 due south. Solution 1. When u=23m/s T=4s A=2m/s2 V=? Therefore we introduce the first equation of motion in this case. V=u + at V=23 + 2(4) V=23 + 8 V=31m/s due north. 2. V=u + at 31=u + 2(4) 31=u + 8