Question

The velocity-time graph for a car is given below. Calculate the total distance and net displacement covered by the car in the entire journey.

• A. $50m,175m$
• B. $225m,125m$
• C. $175m,50m$
• D. $125m,225m$

Answer 1

The correct option is B $225m,125m$

Total area under a velocity time graph gives the net displacement.
Displacement from t = 0 s to t = 25 s
Area of trapezium
= ½ $\times$ (Sum of parallel sides) $\times$ height
= ½ $\times$ (10+25) $\times$ 10
= 175 ${\text{m}}^2$

Displacement from t = 25 s to t = 35 s,
Area of triangle
= ½ $\times$ base $\times$ height
= ½ $\times$ 10 $\times$ -10
= - 50 ${\text{m}}^2$

Net displacement
= Area of trapezium + Area of triangle
= 175 +(- 50)
= 125 m
Net distance = sum of magnitude of displacements
= |Area of trapezium| + |Area of triangle|
= |175| + |-50|
= 225 m