Question 5: Motion in 1D

Question

The velocity $(v)$ versus time $(t)$ plot of a particle is shown in the figure, for a time interval of 40 s . The total distance travelled by the particle and the average velocity during this period are, respectively
$\_\_\_\_$.

Accepted Answer

The displacement is the algebraic sum of the areas (assuming area above the t -axis as positive and area below as negative). And the distance is the sum of the magnitudes of the areas (assuming all areas as positive because distance can't be negative).
The graph consists of two triangles:
For triangle 1 from $\mathrm{t}=0$ to $\mathrm{t}=20 \mathrm{~s}$ :
Base $\mathrm{b}_1=20 \mathrm{~s}$, Height $\mathrm{h}_1=+5 \mathrm{~m} / \mathrm{s}$
So, area of triangle 1 is,
$$ 	ext { Area }_1=\frac{1}{2} 	imes 20 	imes 5=50 \mathrm{~m} $$
For triangle 2 from $\mathrm{t}=20$ to $\mathrm{t}=40 \mathrm{~s}$ :
Base $\mathrm{b}_2=40 \mathrm{~s}-20 \mathrm{~s}=20 \mathrm{~s}$, Height $\mathrm{h}_2=-5 \mathrm{~m} / \mathrm{s}$
So, area of triangle 2 is,
$$ 	ext { Area }_2=\frac{1}{2} 	imes 20 	imes(-5)=-50 \mathrm{~m} $$
Total Distance $=\mid$ Area $_1|+|$ Area $_2|=|50|+|-50|=100 \mathrm{~m}$
The average velocity is defined as the total displacement divided by total time.
$$ \begin{aligned} & 	ext { Total Displacement }=	ext { Area }_1+	ext { Area }_2=50+(-50)=0 \mathrm{~m} \ & 	ext { So, average velocity }=\frac{	ext { Total Displacement }}{	ext { Total Time }}=\frac{0}{40}=0 \mathrm{~m} / \mathrm{s} \end{aligned} $$
Therefore, total distance covered by particle is 100 m and its average velocity is zero.
Hence, the correct option is (C).

Answer

25 m and zero

Answer

50 m and zero

Answer

100 m and zero

Answer

100 m and $2.5 \mathrm{~m} / \mathrm{s}$

Motion in 1D

The velocity (v) versus time (t) plot of a particle is shown in the figure, for a time interval of 40 s . The total distance travelled by the particle and the average velocity during this period are, respectively \_\_\_\_.

Choose the correct answer: