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

(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 of the car?

(a) Distance travelled by the car in the first 4 seconds = area under the slope of the speed-time graph.

To find this area, we will use the square counting method.

On time axis (i.e., x axis) :

5 squares = 2 units

∴ 1 square = $\dfrac{2}{5}$ units

On speed axis (i.e., y axis) :

3 squares = 2 units

∴ 1 square = $\dfrac{2}{3}$ units

Area of each square = $\dfrac{2}{5}$ x $\dfrac{2}{3}$ = $\dfrac{4}{15}$ sq units.

We have fully filled squares, more than half filled squares, half filled squares and less than half filled squares in this area.

Area of fully filled square = Area of 1 square = $\dfrac{4}{15}$ sq units

Area of more than half filled square = Area of 1 square = $\dfrac{4}{15}$ sq units

Area of half filled square = $\dfrac{1}{2}$ x Area of 1 square = $\dfrac{1}{2} \times \dfrac{4}{15}$ = $\dfrac{2}{15}$ sq units

Area of less than half filled square = 0

No. of fully filled squares = 56 [Marked in graph with 'F']

Area of fully filled squares = 56 x $\dfrac{4}{15}$ = 14.93 sq units

No. of more than half filled squares = 4 [Marked in graph with 'M']

Area of more than half filled squares = 4 x $\dfrac{4}{15}$ = 1.06 sq units

No. of half filled squares = 3 [Marked in graph with 'H']

Area of half filled squares = 3 x $\dfrac{4}{15}$ = 0.8 sq units

No. of less than half filled squares = 4 [Marked in graph with 'L']

Area of less than half filled squares = 4 x 0 = 0 sq units

∴ Total Area = 14.93 + 1.06 + 0.4 = 16.39 sq units

∴ Total distance covered in first 4 seconds = Area under speed-time graph = 16.4 m

Hence, the car travels 16.4 m in first 4 seconds

(b) From graph, the speed of the car does not change from the points (x = 6) to (x = 10). Hence, the car is said to be in uniform motion from the 6th to the 10th second.