Physics Question from JEE Main 2023 · 11 April, Shift 1

A particle moves so that its velocity $v$ (in $\text{m/s}$) varies with time $t$ (in $\text{s}$) over the first $25 \, \text{s}$ as follows:

• from $t = 0$ to $t = 5 \, \text{s}$: $v$ rises uniformly from $0$ to $10 \, \text{m/s}$
• from $t = 5 \, \text{s}$ to $t = 10 \, \text{s}$: $v$ stays constant at $10 \, \text{m/s}$
• from $t = 10 \, \text{s}$ to $t = 15 \, \text{s}$: $v$ rises uniformly from $10$ to $20 \, \text{m/s}$
• from $t = 15 \, \text{s}$ to $t = 20 \, \text{s}$: $v$ falls uniformly from $20 \, \text{m/s}$ to $0$
• from $t = 20 \, \text{s}$ to $t = 25 \, \text{s}$: $v$ falls uniformly from $0$ to $-20 \, \text{m/s}$

The ratio of distance to displacement in $25 \, \text{s}$ of motion is: