Motion graphs Archives | School Portal NG

LINEAR MOTION Introduction Study of motion is divided into two; Kinematics Dynamics In kinematics forces causing motion are disregarded while dynamics deals with motion of objects and the forces causing them. Displacement Distance moved by a body in a specified direction is called displacement. It is denoted by letter‘s’ and has both magnitude and direction. Distance is the movement from one point to another. The Si unit for displacement is the metre (m). Speed This is the distance covered per unit time. Speed distance covered/ time taken. Distance is a scalar quantity since it has magnitude only. The SI unit for speed is metres per second(m/s or ms-1) Average speed total distance covered/total time taken Other units for speed used are Km/h. Examples A body covers a distance of 10m in 4 seconds. It rests for 10 seconds and finally covers a distance of 90m in 60 seconds. Calculate the average speed. Solution Total distance covered 10+90 100m Total time taken 4+10+6 20 seconds Therefore average speed 100/20 5m/s Calculate the distance in metres covered by a body moving with a uniform speed of 180 km/h in 30 seconds. Solution Distance coveredspeed*time 180*1000/60*6050m/s 50*30 1,500m Calculate the time in seconds taken a by body moving with a uniform speed of 360km/h to cover a distance of 3,000 km? Solution Speed:360km/h360*1000/60*60100m/s Timedistance/speed 3000*1000/100 30,000 seconds. III. Velocity This is the change of displacement per unit time. It is a vector quantity. Velocitychange in displacement/total time taken The SI units for velocity are m/s Examples A man runs 800m due North in 100 seconds, followed by 400m due South in 80 seconds. Calculate, His average speed His average velocity His change in velocity for the whole journey Solution Average speed: total distance travelled/total time taken 800+400/100+80 1200/180 6.67m/s Average velocity: total displacement/total time 800-400/180 400/180 2.22 m/s due North Change in velocityfinal-initial velocity (800/100)-(400-80) 8-5 3m/s due North A tennis ball hits a vertical wall at a velocity of 10m/s and bounces off at the same velocity. Determine the change in velocity. Solution Initial velocity(u)-10m/s Final velocity (v) 10m/s Therefore change in velocity v-u 10- (-10) 20m/s Acceleration This is the change of velocity per unit time. It is a vector quantity symbolized by ‘a’. Acceleration ‘a’change in velocity/time taken v-u/t The SI units for acceleration are m/s2 Examples 1. The velocity of a body increases from 72 km/h to 144 km/h in 10 seconds. Calculate its acceleration. Solution Initial velocity 72 km/h20m/s Final velocity 144 km/h40m/s Therefore ‘a’ v-u/t 40-20/10 2m/s2 A car is brought to rest from 180km/h in 20 seconds. What is its retardation? Solution Initial velocity180km/h50m/s Final velocity 0 m/s A v-u/t0-50/20 -2.5 m/s2 Hence retardation is 2.5 m/s2 Motion graphs Distance-time graphs Area under velocity-time graph Consider a body with uniform or constant acceleration for time‘t’ seconds; This is equivalent to the area under the graph. The area under velocity-time graph gives the distance covered by the body under‘t’ seconds. Example A car starts from rest and attains a velocity of 72km/h in 10 seconds. It travels at this velocity for 5 seconds and then decelerates to stop after another 6 seconds. Draw a velocity-time graph for this motion. From the graph; Calculate the total distance moved by the car Find the acceleration of the car at each stage. Solution From the graph, total distance covered area of (A+B+C) (1/2×10×20)+(1/2×6×20)+(5×20) 100+60+100 260m Also the area of the trapezium gives the same result. Acceleration gradient of the graph Stage A gradient 20-0/ 10-0 2 m/s2 Stage b gradient 20-20/15-10 0 m/s2 Stage c gradient 0-20/21-15 -3.33 m/s2 Using a ticker-timer to measure speed, velocity and acceleration. It will be noted that the dots pulled at different velocities will be as follows; Most ticker-timers operate at a frequency of 50Hzi.e. 50 cycles per second hence they make 50 dots per second. Time interval between two consecutive dots is given as, 1/50 seconds 0.02 seconds. This time is called a tick. The distance is measured in ten-tick intervals hence time becomes 10×0.02 0.2 seconds. Examples A tape is pulled steadily through a ticker-timer of frequency 50 Hz. Given the outcome below, calculate the velocity with which the tape is pulled. Solution Distance between two consecutive dots 5cm Frequency of the ticker-timer50Hz Time taken between two consecutive dots1/500.02 seconds Therefore, velocity of tape5/0.02 250 cm/s The tape below was produced by a ticker-timer with a frequency of 100Hz. Find the acceleration of the object which was pulling the tape. Solution Time between successive dots1/1000.01 seconds Initial velocity (u) 0.5/0.01 50 cm/s Final velocity (v) 2.5/0.01 250 cm/s Time taken 4 ×0.01 0.04 seconds Therefore, acceleration v-u/t 250-50/0.045,000 cm/s2 Equations of linear motion The following equations are applied for uniformly accelerated motion; v u + at s ut + ½ at2 v2 u2 +2as Examples A body moving with uniform acceleration of 10 m/s2 covers a distance of 320 m. if its initial velocity was 60 m/s. Calculate its final velocity. Solution V2 u2 +2as (60) +2×10×320 3600+6400 10,000 Therefore v (10,000)1/2 v 100m/s A body whose initial velocity is 30 m/s moves with a constant retardation of 3m/s. Calculate the time taken for the body to come to rest. Solution v u+at 0 30-3t 303t t 30 seconds. A body is uniformly accelerated from rest to a final velocityof 100m/s in 10 seconds. Calculate the distance covered. Solution sut+ ½ at2 0×10+ ½ ×10×102 1000/2500m Motion under gravity. Free fall The equations used for constant acceleration can be used to become, v u+gt s ut + ½ gt2 v2 u+2gs Vertical projection Since the body goes against force of gravity then the following equations hold v u-gt ……………1 s ut- ½ gt2 ……2 v2 u-2gs …………3 N.B time taken to reach maximum height is given by the following tu/g since v0 (using equation 1) Time of flight The time taken by the projectile is the timetaken to fall back to its point ofprojection. Using eq. 2 then, displacement 0 0 ut- ½ gt2 02ut-gt2 t(2u-gt)0 Hence, t0 or t 2u/g to corresponds to the start of projection t2u/gcorresponds to the time of flight The time of flight is twice the time taken to attain maximum height. Maximum height reached. Using equation 3 maximum height, Hmax is attained when v0 (final velocity). Hence v2 u2-2gs;- 0u2-2gHmax, therefore 2gHmaxu2 Hmaxu2/2g Velocity to return to point of projection. At the instance of returning to the original point, total displacement equals to zero. v2 u2-2gs hence v2 u2 Thereforevu or v±u Example A stone is projected vertically upwards with a velocity of 30m/s from the ground. Calculate, The time it takes to attain maximum height The time of flight The maximum height reached The velocity with which it lands on the ground. (take g10m/s) Solution Time taken to attain maximum height Tu/g30/103 seconds The time of flight T2t 2×36 seconds Or T2u/g2×30/106 seconds. Maximum height reached Hmax u2/2g 30×30/2×10 45m Velocity of landing (return) v2 u2-2gs, but s0, Hence v2u2 Thereforev(30×30)1/230m/s Horizontal projection The path followed by a body (projectile) is called trajectory. The maximum horizontal distance covered by the projectile is called range. The horizontal displacement ‘R’ at a time‘t’ is given by sut+1/2 at 2 Taking uu and a0 hence Rut, is the horizontal displacement and h1/2gt2 is the vertical displacement. NOTE The time of flight is the same as the time of free fall. Example A ball is thrown from the top of a cliff 20m high with a horizontal velocity of 10m/s. Calculate, The time taken by the ball to strike the ground The distance from the foot of the cliff to where the ball strikes the ground. The vertical velocity at the time it strikes the ground. (take g10m/s) Solution h ½ gt2 20 ½ ×10×t2 4010t2 t240/104 t2 seconds Rut 10×2 20m vu+atgt 2×1020m/s See also: FLUID FLOW SOUND WAVES HOOKES LAW MAGNETIC EFFECT OF AN ELECTRIC CURRENT