The displacement x versus time graph is shown below.
(A) The average velocity during 0 to 3 s is 10 m / s
(B) The average velocity during 3 to 5 s is 0 m / s
(C) The instantaneous velocity at t=2 s is 5 m / s
(D) The average velocity during 5 to 7 s and instantaneous velocity at t=6.5 s are equal
(E) The average velocity from t=0 to t=9 s is zero
Choose the correct answer from the options given below :

Solution

Physics sir · 3 steps

Average velocity joins the end points; instantaneous velocity is the slope of the segment you are on.

1. Read the corners. From the graph: O(0,0), A(1,-5), B(2,0), C(3,5), D(5,5), E(6,10), F(7,0), G(8,-5), H(9,0), with t in s and x in m.
2. Two different rules. Average velocity is Δ x/Δ t between the two end points, whatever happens in between. Instantaneous velocity is the slope of the straight segment at that instant.

   Statement	Calculation	Verdict
   (A) average, 0 to 3 s	(5 - 0)/3 = 5/3 m s⁻¹, not 10	false
   (B) average, 3 to 5 s	(5 - 5)/2 = 0	true
   (C) instantaneous, t = 2 s	slope of AC: (5 - (-5))/(3 - 1) = 5 m s⁻¹	true
   (D) average 5 to 7 s vs instantaneous at 6.5 s	(0 - 5)/2 = -2.5 m s⁻¹ vs slope of EF (0 - 10)/1 = -10 m s⁻¹	false
   (E) average, 0 to 9 s	(0 - 0)/9 = 0	true

3. So. Statements (B), (C) and (E) are correct, which is option (d).

The idea to keep. Average velocity looks only at the start and end points: Δ x/Δ t. Instantaneous velocity is the slope of the straight piece at that moment. They match only when the interval lies on one straight segment.

Answer from the NTA final answer key. Solution by PrepWiser.