Physics · Motion In A Straight Line · NEET
It gives displacement. Displacement = velocity x time, and on a v-t graph velocity x time is exactly the area between the line and the time axis. To get distance instead, you must add up the sizes of all areas as positive numbers (ignore the minus sign for parts below the axis).
Area below the time axis means velocity is negative, so the object is moving backward (in the negative direction). That area is counted as a negative displacement. If the object first moves forward (area above) and then backward (area below) by the same amount, the two areas cancel and net displacement is zero, even though the object did move.
Split the graph into simple shapes: rectangles (constant velocity), triangles (uniform acceleration from or to zero), and trapeziums. Find each shape's area using area = base x height for a rectangle, or area = 1/2 x base x height for a triangle. Give areas above the axis a plus sign and areas below a minus sign. Add them all up. The total is the displacement.
For a tiny time step, distance moved = velocity x that small time = a thin strip of area under the graph. Adding all the thin strips over the whole interval adds all the small displacements, which sums to the total displacement. NCERT states this directly: the area under the v-t curve between two times equals the displacement in that interval.
Yes. The area rule is always true, curved or straight. For a straight-line v-t graph you use simple triangle and rectangle formulas. For a curved graph the exact area needs calculus (integration), but the idea is the same: area = displacement.
A toy car with charge q moves on a frictionless horizontal plane under a uniform electric field E. Due to the force qE, its velocity increases from 0 to 6 m/s in one second. At that instant the direction of the field is reversed. The car moves for two more seconds under this field. The average velocity and the average speed of the toy car are respectively:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Yes, the signed area (above positive, below negative) between the v-t line and the time axis is always the displacement, for both straight and curved graphs.
Take the size of every area piece as a positive number and add them. Never let a below-axis area subtract when you want distance.
Rectangle area = base x height (constant velocity). Triangle area = 1/2 x base x height (velocity rising or falling to zero). Trapezium area = 1/2 x (sum of two parallel sides) x height.
Yes. If the object goes forward then back, the positive area above the axis and the negative area below can cancel to give zero displacement, while the distance (sum of sizes) is not zero.
NEET regularly asks average velocity vs average speed using v-t graphs where the field or acceleration reverses (like the 2018 toy-car question). The area rule with correct signs is the fastest way to solve them without kinematic equations.
