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# The speed-time graph for a car is shown in Fig. 8.12.

**Fig. 8.12**

$(a)$. Find how far does the car travel in the first 4 seconds.Shade the area on the graph that represents the distance travelled by the car during the period.

$(b)$. Which part of the graph represents uniform motion ofthe car?"

(a)

__In a velocity-time graph__

Distance = Area of v-t graph.

Here, we find the approximate area using Area of triangle (shaded region in sky-blue colour).

Distance = Area of △OAB

= $\frac{1}{2}\times b\times h$, where b = base and h = height

= $\frac{1}{2}\times 4\times 6$ $[\because b=time,\ h=speed]$

= $12m$

We still have not the total area, since, the portion shaded with yellow colour is not covered. So, this is not accurate.

Therefore, to find the area of the yellow region along with the sky-blue region, we need to find the area by counting the squares.

From the graph, we can see that-

Number of squares on the horizontal axis (time axis)

5 squares = 2 units.

$1\ square=\frac{2}{5}units$

Number of squares on the vertical axis (speed axis)

3 squares = 2 units.

$1\ square=\frac{2}{3}units$

$\therefore Area\ of\ each\ square=\frac{2}{5}\times \frac{2}{3}\Leftrightarrow \frac{4}{5}sq\ units$

So, 1 square represents $\frac{4}{15}m$ distance.

Now, we take

Symbol | Nature of shaded region | Number | Area |

Square | Full Square | 57 | $57\times \frac{4}{15}=15.2$ |

Triangle | More than half-square | 3 | $3\times \frac{4}{15}=0.8$ |

Star | Half-square | 3 | $3\times \frac{1}{2}\times \frac{4}{15}=0.4$ |

Circle | Less than half-square | 4 | $4\times 0=0$ |

$\therefore Total\ area=15.2+0.8+0.4+0\Leftrightarrow 16.4sq\ units$

Hence, the distance covered by the car in 0-4 sec = 16.4m

(b) The part of the graph in green colour between time, t = 6 s to 10 s represents the uniform motion of the car, because, between this time, the speed of the car becomes constant.

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