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CBSE Class 11-science Answered

Question 2, 3, 4
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Asked by Sritam.sen.09 | 27 Sep, 2018, 10:42: PM
answered-by-expert Expert Answer
Only one question per posting will be answered. Hence only Question No. 2 is answered
 
Q.2.(a)
Let from time t1 to t2, an object travels with constant veleocity v. This is plotted in the velocity-time graph shown in above figure.
 
Let t is time duration between the instants of time  t1 and t2.
 
Area of the rectangle = AD×AB = Velocity × (time-interval) = v×(t2-t1) = v×t = distance travelled
 
Hence area under the velocity-time curve is the distance travelled.
 
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Q.2(b)
Let an object travel from instance of time t0 to that of t1 with uniform acceleration.
 
Let the initial velocity is v0 and final velocity is v1. This is plotted in the velocity-time graph shown in above figure.
 
We know that area under the curve of velocity-time graph is the distance travelled.
Hence the distance travelled from t0 to t1 is the area of the trapezium ABCD
 
Area of trapezium = (1/2)×(AD+BC)×AB = (1/2)×(v0+v1)×(t1-t0) ..................(1)
 
Let t is the time duration between the instances of time t0 and t1.
when velocity increases with uniform acceleration a in a time interval t, we have v1 = v0+a×t .................(2)
 
substituting for v1 using eqn.(2) in eqn.(1), we have,
 
Area of trapezium = distance travelled S = v0×t + (1/2)×a×t2
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Q.2(c)
 
To get retardation, we need to use the equation " v2 = u2 -2×a×S ",
where u and v are initial and final velocity respectively when the object travelled a distance S.
 
when the moving object retarded and comes to rest, we have v =0 .
 
initial speed = 126 km/h = 126×(5/18) = 35 m/s
 
henced retardation, a = u2 / (2×S) = (35×35) /(2×200)  = 3.0625 m/s2
Answered by Thiyagarajan K | 28 Sep, 2018, 08:58: AM
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