Motion in a Straight Line – Graphs Related to Motion: NEET MCQs with Answers

Explanation: Instantaneous velocity is the slope of the distance-time curve. The curve is steepest around point C. The slope is small near A and B and has decreased again by D.

Explanation: A particle in one-dimensional motion can have only one velocity at any instant, so velocity must be a single-valued function of time. On a v–t graph, any vertical line may therefore cut the curve at only one point. The closed circle fails this test: for one value of t it gives two different values of v, which would mean the particle has two velocities at the same instant. That is impossible. It would also need time to run backwards along part of the loop. The triangular graph, the rising curve and the sine-like curve each give a single velocity at every instant, so all of them are acceptable. In the sine-like curve the velocity simply reverses direction periodically.

Explanation: Instantaneous velocity is the slope of the displacement–time graph. It is negative wherever the curve is falling, i.e. where displacement is decreasing with time. Of the four marked points, the curve is clearly descending only at E, so the slope — and hence the instantaneous velocity — is negative there. At C the curve is rising, so the velocity is positive; at D the curve is near a local maximum, where the slope is approximately zero; and at F the curve is rising steeply again, giving a positive velocity.

Explanation: The instantaneous velocity at any instant is the slope of the tangent drawn to the distance–time curve at that instant, v = dx/dt. The velocity is therefore greatest where the curve is steepest. Reading the graph, the curve rises gently near A and B, becomes steepest around C, and then flattens out again towards D. Hence the maximum instantaneous velocity occurs around the point C. Recall the general shapes: a horizontal line means the object is at rest, a straight sloping line means uniform velocity, and a curve means the velocity is changing.

Explanation: Distance = area under the s-t (speed-time) graph = area of the triangle = ½ × base × height = ½ × 8 × 20 = 80 m.

Explanation: Positive, increasing acceleration means the velocity (the slope of the x-t curve) keeps increasing — this appears as a curve that gets progressively steeper, i.e., concave upward throughout.

Explanation: On a displacement–time graph the slope gives velocity: v ∝ tanθ. Ratio = tan30° : tan45° = (1/√3) : 1 = 1 : √3.

Explanation: Instantaneous velocity is the slope of the tangent to the position–time curve. At the highest point of a smooth curve the tangent is horizontal, so the slope is zero. The car is momentarily at rest as it turns back.

Explanation: The slope of a position–time graph is the velocity. Positive acceleration means the velocity increases, so the slope must keep increasing and the curve bends upward. Straight lines show constant velocity and a downward-bending curve shows negative acceleration.

Explanation: The three stages give displacements of +6 m, −4 m and +6 m. Distance adds their sizes: 6 + 4 + 6 = 16 m. Displacement adds them with sign: 6 − 4 + 6 = 8 m.

Explanation: Constant positive acceleration makes velocity rise linearly, with slope equal to the acceleration. Zero acceleration keeps the velocity constant, giving a horizontal segment. When the same acceleration returns, the velocity again rises with the same slope.

Explanation: The change in velocity equals the area under the acceleration–time graph. That area is ½ × 4 × 12 = 24 m/s for the first 4 s plus 4 × 12 = 48 m/s for the next 4 s, a total of 72 m/s. Average acceleration = 72/8 = 9 m/s².

Explanation: Distance travelled is a running total of path length, so it can stay the same or increase but can never decrease with time. A falling line would mean the distance covered gets smaller, which is impossible. The other shapes show uniform motion, accelerated motion and a body that stops.

Explanation: A straight line on a position–time graph means constant velocity, equal to its slope. P has a slope of 3 m/s and Q a slope of 1 m/s, so P is faster. The starting position only shifts the line and does not affect the velocity.

Explanation: The slope of the graph is −2 m/s² throughout, so the acceleration never changes sign. The velocity becomes zero at t = 6 s and then turns negative, so the particle reverses. The areas from 0 to 6 s and from 6 s to 12 s are each 36 m in size but opposite in sign, giving zero net displacement and an average speed of 6 m/s in each half.