Question

The velocity time graph for a particle is shown in figure, then

What is the distance and displacement of the particle from 0 to 15 seconds.

Answer 1

Displacement = area under the velocity time graph = ar(oab)+ ar(under bc) =
$(1 \div 2) \times 10 \times 15 + 5 \times 5 = 100$
here the motion is a straight line, therefore
Distance = Displacement = 100

Answer 2

Answer:

The distance and displacement of the particle from 0 to 15 seconds is 100m and 100m respectively.

Explanation:

The displacement of a particle is equal to the area under the velocity-time graph.

Dividing the graph into two geometries, one of triangle and the other a square. The displacement will be sum of the area of triangle and square.

Area of triangle is given by

$\dfrac{1}{2} \times base \times height$

From the graph, area under triangle is

$\dfrac{1}{2} \times 10 \times 15=75m^{2}$

Area of square is given by $side \times side$

From the graph, area under square is $5 \times 5=25m^{2}$

Therefore, the displacement of the particle is $75+25=100m$.

Since the particle is moving with constant acceleration, displacement is equal to the distance travelled, therefore the distance travelled is 100m.

Hence, the distance and displacement of the particle from 0 to 15 seconds is 100m and 100m respectively.