derive all three equations of uniform motion accelerated motion by graphical method
Answers
Answered by
3
The three equations of motion v = u + at ; s = ut + (1/2) at2 and v2 = u2 + 2as can be derived with the help of graphs as described below.1. Derive v = u + at by Graphical MethodConsider the velocity – time graph of a body shown in the below Figure.Velocity–Time graph to derive the equations of motion.The body has an initial velocity u at point A and then its velocity changes at a uniform rate from A to B in time t. In other words, there is a uniform acceleration 'a' from Ato B, and after time t its final velocity becomes 'v' which is equal to BC in the graph. The time t is represented by OC. To complete the figure, we draw the perpendicular CB from point C, and draw AD parallel to OC. BE is the perpendicular from point B to OE.Now, Initial velocity of the body, u=OA...... (1)And, Final velocity of the body, v=BC........ (2)But from the graph BC=BD + DCTherefore, v=BD + DC ......... (3)Again DC=OASo, v=BD + OANow, From equation (1), OA=uSo, v=BD + u ........... (4)We should find out the value of BD now. We know that the slope of a velocity – time graph is equal to acceleration, a.Thus, Acceleration, a=slope of line ABor a=BD/ADBut AD=OC = t,so putting t in place of AD in the above relation, we get:a=BD/tor BD=atNow, putting this value of BD in equation (4) we get :v=at + uThis equation can be rearranged to give:v=u + atAnd this is the first equation of motion. It has been derived here by the graphical method.2. Derive s = ut + (1/2) at2 by Graphical MethodVelocity–Time graph to derive the equations of motion.Suppose the body travels a distance s in time t. In the above Figure, the distance travelled by the body is given by the area of the space between the velocity – time graph AB and the time axis OC,which is equal to the area of the figure OABC. Thus:Distance travelled=Area of figure OABC=Area of rectangle OADC + Area of triangle ABDWe will now find out the area of the rectangle OADC and the area of the triangle ABD.(i) Area of rectangle OADC=OA × OC=u × t=ut ...... (5)(ii) Area of triangle ABD=(1/2) × Area of rectangle AEBD=(1/2) × AD × BD=(1/2) × t × at (because AD = t and BD = at)=(1/2) at2...... (6)So, Distance travelled, s=Area of rectangle OADC + Area of triangle ABDor s = ut + (1/2) at2
This is the second equation of motion. It has been derived here by the graphical method.3. Derive v2 = u2 + 2as by Graphical MethodVelocity–Time graph to derive the equations of motion.We have just seen that the distance travelled s by a body in time t is given by the area of the figure OABC which is a trapezium. In other words,Distance travelled, s=Area of trapezium OABCNow, OA + CB = u + v and OC = t. Putting these values in the above relation, we get: ...... (7)We now want to eliminate t from the above equation. This can be done by obtaining the value of t from the first equation of motion. Thus, v = u + at (First equation of motion)
And, at = v – u or Now, putting this value of t in equation (7) above, we get: or 2as=v2 – u2 [because (v + u) × (v – u) = v2 – u2]or v2=u2 + 2asThis is the third equation of motion.
This is the second equation of motion. It has been derived here by the graphical method.3. Derive v2 = u2 + 2as by Graphical MethodVelocity–Time graph to derive the equations of motion.We have just seen that the distance travelled s by a body in time t is given by the area of the figure OABC which is a trapezium. In other words,Distance travelled, s=Area of trapezium OABCNow, OA + CB = u + v and OC = t. Putting these values in the above relation, we get: ...... (7)We now want to eliminate t from the above equation. This can be done by obtaining the value of t from the first equation of motion. Thus, v = u + at (First equation of motion)
And, at = v – u or Now, putting this value of t in equation (7) above, we get: or 2as=v2 – u2 [because (v + u) × (v – u) = v2 – u2]or v2=u2 + 2asThis is the third equation of motion.
Similar questions