Question 50: Solution
1. Geometric Interpretation of Distance Loss
The distance covered is the area under the Velocity-Time graph. We are comparing the area of the original curve against the area of the curve with the gear shift.
The “gear shift curve” is identical to the original curve but shifted to the right by $\tau = 1.0\,\text{s}$ for the duration after the shift event. The loss in distance is represented by the area of the “strip” between these two curves.
2. Calculation
The “missing area” corresponds to the vertical strip created by the horizontal time shift. The width of the shift is $\tau = 1.0 \, \text{s}$. The effective height of this strip spans from the velocity at the shift ($v_{shift}$) to the final velocity ($v_{final}$).
Using the values derived in Q49:
- Shift Velocity ($v_{shift}$) = $25 \, \text{m/s}$
- Final Terminal Velocity ($v_{final}$) = $40 \, \text{m/s}$
- Delay ($\tau$) = $1.0 \, \text{s}$
$$ \Delta S \approx \tau (v_{final} – v_{shift}) $$
$$ \Delta S = 1.0 \times (40 – 25) $$
$$ \Delta S = 15 \, \text{m} $$
Correct Option: (b) 15 m