A body starts from rest and moves with a uniform acceleration of 5m/s2 for 5s and then it moves with a constant velocity for 4s. Later it slows down and comes to rest in 5s. Draw the velocity time graph for the motion of the body a) What is the maximum velocity attained by the body?b) What is the distance travelled during this period of acceleration?c) What is distance travelled when the body was moving with constant velocity?d) What is the retardation of the body while slowing down?e) What is the distance travelled by retarding?f) What is the total distance travelled?
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See the diagram enclosed.
At t=0, start at O, origin. Draw a line to A with a slope of5 for 5 seconds. For every one second move up by5 units vertically.Then draw a horizontal line for 4 seconds reaching
D. here t= 9sec.
Now join D with E at t= 9+5 = 14sec on time sline when the body stops
ie., velocity is 0. The slope here is found by height lost / horizontal span = 25/5 = 5 m/sec²
a) V at A : 25 m/sec
b) Area enclosed bytriangle OAB = 1/2* 25 * 5 = 62.5 m
c) area of ABCD = 25 * 4 = 100 m
d ) slope = (0-25)/5= -5 m/sec²
e ) area of triangle DCE = 62.5 mf ) total area bounded between the OADE and time axis. = 62.5 + 100 + 62.5 = 230m
At t=0, start at O, origin. Draw a line to A with a slope of5 for 5 seconds. For every one second move up by5 units vertically.Then draw a horizontal line for 4 seconds reaching
D. here t= 9sec.
Now join D with E at t= 9+5 = 14sec on time sline when the body stops
ie., velocity is 0. The slope here is found by height lost / horizontal span = 25/5 = 5 m/sec²
a) V at A : 25 m/sec
b) Area enclosed bytriangle OAB = 1/2* 25 * 5 = 62.5 m
c) area of ABCD = 25 * 4 = 100 m
d ) slope = (0-25)/5= -5 m/sec²
e ) area of triangle DCE = 62.5 mf ) total area bounded between the OADE and time axis. = 62.5 + 100 + 62.5 = 230m
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