Figure shows the velocity-time graph of a particle moving in a straight line.

(i) State the nature of motion of particle.

(ii) Find the displacement of particle at t = 6 s.

(iii) Does the particle change its direction of motion?

(iv) Compare the distance travelled by the particle from 0 to 4 s and from 4 s to 6 s.

(v) Find the acceleration from 0 to 4 s and retardation from 4 s to 6 s.

(i) As we observe the graph, we find that, the nature of motion of particle is that, the particles are uniformly accelerated from 0 to 4s and then uniformly retarded from 4s to 6s.

(ii) As we know,

displacement of particles can be obtained by finding the area enclosed by the graph in that part with the time axis up to that instance.

At t = 6 s,

Displacement = area of triangle

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

Substituting the values in the formula above, we get,

$= \dfrac {1}{2} \times {(6 - 0) \text {s}} \times (2 - 0)\text {m s}^{-1} \\[0.5em] = \dfrac {1}{2} \times 6 \times 2 \text {m} \\[0.5em] = 6 \text {m} \\[0.5em]$

Hence,
Displacement of particle at t = 6 s is 6 m

(iii) No, the particle does not change its direction of motion.

(iv) At t = 0 to 4 s,

Distance covered = area of triangle

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

Substituting the values in the formula above, we get,

$= \dfrac {1}{2} \times {(4 - 0) \text {s}} \times (2 - 0)\text {m s}^{-1} \\[0.5em] = \dfrac {1}{2} \times 4 \text {s} \times 2 \text {m s}^{-1} \\[0.5em] = 4 \text {m} \\[0.5em]$

Hence,
Distance covered between 0 to 4 s = 4 m

At t = 4 s to 6 s,

Distance covered = area of triangle

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

Substituting the values in the formula above, we get,

$= \dfrac {1}{2} \times {(6 - 4) \text {s}} \times (2 - 0)\text {m s}^{-1} \\[0.5em] = \dfrac {1}{2} \times 2 \text {s} \times 2 \text {m s}^{-1} \\[0.5em] = 2 \text {m} \\[0.5em]$

Hence,
Distance covered between 4s to 6 s = 2 m

∴ Distance covered between 0 to 4 s : Distance covered between 4 s to 6 s = 4 : 2 = 2 : 1

(v) Acceleration in part 0 s to 4 s = slope of graph

$\text {slope} = \dfrac{(2 - 0) \text {ms}^{-1}}{(4-0) \text {s}} \\[0.5em] \text {slope} = \dfrac{2 \text {ms}^{-1}}{4 \text {s}} \\[0.5em] \text {slope} = 0.5 \text {ms}^{-2} \\[0.5em]$

Hence,
Acceleration in part 0 s to 4 s = 0.5 m s^-2

As we know,

Retardation in part 4 s to 6 s = slope of graph

$\text {slope} = \dfrac{(0 - 2) \text {ms}^{-1}}{(6 - 4) \text {s}} \\[0.5em] \text {slope} = \dfrac{- 2 \text {ms}^{-1}}{2\text {s}} \\[0.5em] \text {slope} = - 1 \text {ms}^{-2} \\[0.5em]$

Acceleration = -1 ms^-2 and as we know retardation is negative acceleration

Hence, Retardation in part 4 s to 6 s = 1 m s^-2