Question

How to solve these questions?

Answer 1

(i) We know the displacement of a velocity - time graph is given by the area under the graph.

So the displacement for OA is equal to the area under the portion OA (that leftmost triangle).

$\sf{\longrightarrow OA=\dfrac{1}{2}(10-0)(20-0)\ m}$

$\sf{\longrightarrow\underline{\underline{OA=100\ m}}}$

The displacement for AB is equal to the area under the portion AB (that middle rectangle).

$\sf{\longrightarrow AB=(20-10)(20-0)\ m}$

$\sf{\longrightarrow\underline{\underline{AB=200\ m}}}$

And the displacement for BC is equal to the area under the portion AB (that rightmost triangle).

$\sf{\longrightarrow BC=\dfrac{1}{2}(25-20)(20-0)\ m}$

$\sf{\longrightarrow\underline{\underline{BC=50\ m}}}$

(ii) We know acceleration of a velocity - time graph at a particular interval is given by the slope of the graph in that interval.

Let us write coordinates for each point in the graph.

$\sf{\longrightarrow O=(0,\ 0)}$

$\sf{\longrightarrow A=(10,\ 20)}$

$\sf{\longrightarrow B=(20,\ 20)}$

$\sf{\longrightarrow C=(25,\ 0)}$

Then, acceleration during OA,

$\sf{\longrightarrow OA=\dfrac{20-0}{10-0}}$

$\sf{\longrightarrow \underline{\underline{OA=2}}}$

Acceleration during AB,

$\sf{\longrightarrow AB=\dfrac{20-20}{20-10}}$

$\sf{\longrightarrow\underline{\underline{AB=0}}}$

Acceleration during BC,

$\sf{\longrightarrow AB=\dfrac{0-20}{25-20}}$

$\sf{\longrightarrow\underline{\underline{AB=-4}}}$

(iii) Average velocity is equal to the total displacement divided by total interval of time.

Total displacement $\sf{=100+200+50=350\ m.}$

Total time $\sf{=25\ s.}$

Then, average velocity,

$\sf{\longrightarrow v_{av}=\dfrac{350\ m}{25\ s}}$

$\sf{\longrightarrow\underline{\underline{v_{av}=14\ m\,s^{-1}}}}$

(iv) Maximum velocity is the maximum possible value of velocity (y value) through which the graph passes.

Here we see the maximum value of velocity is $\sf{20\ m\,s^{-1}.}$

$\sf{\longrightarrow\underline{\underline{v_{max}=20\ m\,s^{-1}}}}$