Question

Fig. 2.41
Fig. 2.42 given below shows a velocity-time graph
for a car starting from rest. The graph has three parts
AB, BC and CD.
(ii) Compare the distance travelled in part BC with
the distance travelled in part AB.​

Answer 1

Answer:

The distance traveled in part BC is twice the distance traveled in part AB.

Explanation:

The velocity time graph can give us the distance traveled if we find the area under the graph.

To find the area under the curve for part AB:

We can clearly see that it is a right angled triangle.

Base of this triangle can be calculated from x axis where time is plotted.

Base of triangle = (t-0) = t units

Height of triangle can be calculated from y axis where velocity is plotted.

Height of triangle = ($v_0 -0$) = $v_0$ units

Area of a triangle is given as:

$A = \dfrac{1}{2} \times \text{Base} \times \text{Height}\\\Rightarrow \text{Area during part AB} = \dfrac{1}{2} \times t \times v_0 \\\Rightarrow \text{Distance traveled during part AB} = \dfrac{1}{2} \times t \times v_0 ...... (1)$

During part BC:

The graph is a rectangle here,

With length towards x axis where time is plotted.

with width towards y axis where velocity is plotted.

Length = (2t - t) = t units

Width = ($v_0 -0$) = $v_0$ units

Area of rectangle represents the distance traveled during part BC.

Distance traveled during part BC = $t \times v_0 ....... (2)$

Comparing equations (1) and (2):

Distance traveled during part BC is twice the distance traveled during part AB.