A scooter accelerates from rest for time t1 at constant rate a1 and then retards at constant rate a2

Q: Why do we use a negative sign for $a_2$ in the second equation?

The problem states that the scooter 'retards' at a rate $a_2$. Retardation means the velocity is decreasing, which implies the acceleration vector is pointing opposite to the velocity vector. In our standard sign convention, if velocity is positive, retardation must be substituted as a negative acceleration ($-a_2$) in the equations of motion.

Q: Could we have solved this using the $v-t$ graph directly without equations?

Absolutely! In a $v-t$ graph, the slope of the line represents acceleration. The slope of the first line is $\frac{v_{\max}}{t_1} = a_1$. The magnitude of the slope of the second line is $\frac{v_{\max}}{t_2} = a_2$. From these, $v_{\max} = a_1 t_1$ and $v_{\max} = a_2 t_2$. Equating them gives the same result instantly.

Q: What if the initial velocity of the scooter wasn't zero?

If the initial velocity was $u$, the first equation would be $v_{\max} = u + a_1 t_1$. The second equation would still be $0 = v_{\max} - a_2 t_2$. Equating them would give $u + a_1 t_1 = a_2 t_2$, and the simple ratio $\frac{t_1}{t_2}$ would no longer just be $\frac{a_2}{a_1}$. It would depend on $u$ as well.

Q: Is the total time of the journey $t_1 + t_2$?

Yes, the total time elapsed from the moment the scooter starts moving to the moment it comes to a complete halt is exactly $t_1 + t_2$.

Q: How would we find the total distance traveled?

The total distance is the area under the $v-t$ graph. Since the graph forms a triangle with base $(t_1 + t_2)$ and height $v_{\max}$, the total distance is $\frac{1}{2} \times (t_1 + t_2) \times v_{\max}$. You can substitute $v_{\max} = a_1 t_1$ or $a_2 t_2$ to get the answer in terms of given variables.

Question

A scooter accelerates from rest for time $t_1$ at constant rate $a_1$ and then retards at constant rate $a_2$ for time $t_2$ and comes to rest. The correct value of $\frac{t_1}{t_2}$ will be

Options:

(A) $\frac{a_1 + a_2}{a_2}$

(B) $\frac{a_2}{a_1}$

(C) $\frac{a_1}{a_2}$

(D) $\frac{a_1 + a_2}{a_1}$

Sigma Percentile · Accepted Answer

### Visualizing the Journey

Imagine you are riding a scooter. You start from a red light, twist the throttle, and speed up. After a while, you see another red light ahead, so you apply the brakes and come to a smooth stop. This everyday scenario is exactly what this physics problem is about!

To solve this elegantly, we can use a powerful tool: the **Velocity-Time ($v-t$) graph**.

When a body starts from rest and accelerates uniformly, its $v-t$ graph is a straight line starting from the origin with a positive slope. When it retards uniformly to rest, the graph is a straight line sloping downwards to the time axis. The peak of this "triangle" represents the maximum velocity, $v_{\max}$, achieved during the journey.

### Phase 1: The Acceleration

Let's break the journey into two parts. In the first part, the scooter starts from rest ($u = 0$) and accelerates at a constant rate $a_1$ for a time $t_1$. At the end of this time, it reaches its peak speed, $v_{\max}$.

We can relate these quantities using the first equation of motion:
$$v = u + at$$

Substituting our values for the first phase:
$$v_{\max} = 0 + a_1 t_1$$
$$v_{\max} = a_1 t_1 \quad \dots (i)$$

This equation tells us exactly how fast the scooter was going right before the brakes were applied.

### Phase 2: The Retardation

Now, the scooter is cruising at $v_{\max}$ and the brakes are hit. This is our new initial velocity for the second phase. The scooter slows down (retards) at a rate of $a_2$. In physics, retardation is just negative acceleration, so we use $-a_2$. It takes a time $t_2$ to finally come to a stop ($v = 0$).

Let's apply the first equation of motion again to this braking phase:
$$0 = v_{\max} - a_2 t_2$$

Rearranging this to solve for $v_{\max}$ gives us:
$$v_{\max} = a_2 t_2 \quad \dots (ii)$$

### The Grand Unification

Look at equations (i) and (ii). Both of them give us an expression for the exact same physical quantity: the maximum velocity $v_{\max}$. Because they represent the same thing, we can set them equal to each other!

$$a_1 t_1 = a_2 t_2$$

We are looking for the ratio of the times, $\frac{t_1}{t_2}$. A simple cross-multiplication gives us our final, elegant result:

$$\frac{t_1}{t_2} = \frac{a_2}{a_1}$$

**Pro Tip:** Notice how the ratio of times is the *inverse* ratio of the accelerations. If you brake twice as hard as you accelerated, it will take you half the time to stop! This intuitive understanding is what makes physics so beautiful.

Animated Solution for Physics - Kinematics: A scooter accelerates from rest for time $t_{1}$ at constant rate $a_{1}$ and then retards at constant rate $a_{2}$ for time $t_{2}$ and comes to rest. The correct value of $\frac{t _{1}}{t _{2}}$ will be

Select Answer:

Visualized Solution

Visualizing the Motion

The First Equation of Motion

Phase 1: Acceleration

Equation for Phase 1

Phase 2: Retardation

Equation for Phase 2

Finding the Ratio

The Way Forward

The Sigma Insight: Motion in a Straight Line

Visualizing the Journey

Phase 1

Phase 2

The Grand Unification

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