MOTION AND THE HUMAN BODY

Motion and the Human Body



Motion and the Human Body

Where necessary in this assignment, take acceleration due to gravity g = 9.81 ms-2.

The graph below shows the displacement of a child walking to school.

The time rate of change of position of a body in a particular direction. Linear velocity is velocity along a straight line, and its magnitude is commonly measured in such units as meters per second (m/s), feet per second (ft/s), and miles per hour (mi/h). Since both a magnitude and a direction are implied in a measurement of velocity, velocity is a directed or vector quantity, and to specify a velocity completely, the direction must always be given. The magnitude only is called the speed.

A body need not move in a straight line path to possess linear velocity. When a body is constrained to move along a curved path, it possesses at any point an instantaneous linear velocity in the direction of the tangent to the curve at that point. The average value of the linear velocity is defined as the ratio of the displacement to the elapsed time interval during which the displacement took place.

2.Athlete A, competing in a 100 m race, crosses the finish line in a time of 10.2 s. At the start, the athlete accelerates uniformly to a top speed in 2.0 s and then remains at a constant speed for the remainder of the race.

(a)Calculate

(i)the average speed of the athlete over the full distance,

1) 100/10.2 = 9.8m/s

(ii)the maximum speed of the athlete if the acceleration were 5.4 m s-2,

v= u + at= 0 + (5.4 x 2)= 10.8m/s

(iii)the distance travelled by the athlete whilst accelerating.

s= ut + (a(t^2))/2= 0 x 2 + (5.4(2^2))/2= 0 + (5.4 x 4)/2= 21.6/2= 10.8m

(b)The graph is a speed time graph for athlete B in the same race.

Using the same axes, draw a speed time graph for athlete A.

(c)Some time after the start of the race the two athletes are running at the same speed. Use your graph to determine

(i)the time at which this occurs,

(ii)the distance covered by the athletes up to this time,

Athlete A:

Athlete B:

(iii)how far apart the athletes are at this time.

A hiker traveled 80.0 m [S] at 1.00 m/s, then 80.0 m [S] at 5.00 m/s. What is the hiker's average velocity?displacement = 160.0 m [S]time for the first part is 80.0÷1.00= 80.0 s, time for the second part is 80.0 m ÷ 5.00 m/s = 16.0 s. Total time = 80.0+16.0 = 96.0 sTherefore, the velocity is (160.0 m [S])÷96.0 s = 1.67 m/s [S]

(b)On the axes below sketch the graphs for speed, v, and distance travelled, s, against time, t, for the first 8.0 s of the car's motion.

4.A mass of 1500kg is attached to a cable and raised vertically by a crane. The graph shows how its velocity varies with time.

(i)(use of v2 = u2 + 2as gives) 322 = (0) + 2 × ...