ncert mod qn-3.28

3.28 The velocity-time graph of a particle in one-dimensional motion is shown in
Fig. 3.29 :

Which of the following formulae are correct for describing the motion of the particle
over the time-interval t1to t2:

(a) x(t2 ) = x(t1) + v (t1) (t2 – t1) +(½) a (t2 – t1)2
(b) v(t2 ) = v(t1) + a (t2 – t1)
(c) vaverage = (x(t2) – x(t1))/(t2 – t1)

(d) aaverage = (v(t2) – v(t1))/(t2 – t1)

(e) x(t2 ) = x(t1) + vaverage (t2 – t1) + (½) aaverage (t2 – t1)2

(f) x(t2 ) – x(t1) = area under the v-t curve bounded by the t-axis and the dotted line
shown.

pls explain why the options a,b and e are ruled out.

5 Answers

1
greatvishal swami ·

bcoz the acc is nt const

21
amit sahoo ·

the graph is not a straight line so acceleration is not constant and thus we cannot use instantaneous value of acceleration which are being used in options a and b.

1
paromita89 paromita ·

wat abt e ?

1357
Manish Shankar ·

in(e)
x(t2)-x(t1) - (1/2)aaverage(t2-t1)2=Vaverage(t2-t1)

Vaverage=[x(t2)-x(t1)]/(t2-t1) -(1/2)aaverage(t2-t1)

But by definition Vaverage=[x(t2)-x(t1)]/(t2-t1) only

thats why e is ruled out

1
paromita89 paromita ·

ok sir..

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