Question

The velocity-time ($v-t$) graph of a body is shown. For the intervals OC and CB, the ratio of the distances covered is $x:y$. Find $x+y$.

Answer 1

4

Answer 2

Solution:

1. Let the coordinates be:
   O = (0, 0), A = ($t_a$, $v_a$) with $v_a$ = h, C = ($t_a$, 0) and B = ($t_b$, 0).

2. From the angle at O:
   tan 30° = $v_a$/$t_a$ = 1/√3 ⟹ $v_a$ = $t_a$/√3 ⟹ $t_a$ = √3 · $v_a$.

3. From the angle at B on segment AB:
   tan 60° = $v_a$/($t_b$ – $t_a$) = √3 ⟹ $t_b$ – $t_a$ = $v_a$/√3.

4. Distance covered in OC (from 0 to $t_a$):
   Area of triangle = ½ · $t_a$ · $v_a$.

5. Distance covered in CB (from $t_a$ to $t_b$):
   Area of triangle = ½ · ($t_b$ – $t_a$) · $v_a$.

6. Ratio of distances = (½ · $t_a$ · $v_a$) : (½ · ($t_b$ – $t_a$) · $v_a$) = $t_a$ : ($t_b$ – $t_a$).
   Substitute $t_a$ = √3 · $v_a$ and $t_b$ – $t_a$ = $v_a$/√3:
   Ratio = (√3 · $v_a$) : ($v_a$/√3) = √3 : (1/√3) = 3:1, i.e., x : y = 3 : 1.

7. Thus, x + y = 3 + 1 = 4.