Q.Figure 2.11 shows the $x$-$t$ plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for $t < 0$ and on a parabolic path for $t > 0$? If not, suggest a suitable physical context for this graph.

Concept and Intuition

When time is expressed as a function of position rather than the usual $x(t)$, we invert the roles: velocity is the reciprocal of $dt/dx$, and acceleration must be found via the chain rule $a=\dfrac{dv}{dt}=\dfrac{dv}{dx}\cdot\dfrac{dx}{dt}=\dfrac{dv}{dx}\cdot v$.

Step-by-Step Solution

1. $t=Px^2+Qx\Rightarrow \dfrac{dt}{dx}=2Px+Q$.
2. Velocity: $v=\dfrac{dx}{dt}=\dfrac{1}{2Px+Q}=(2Px+Q)^{-1}$.
3. Differentiate $v$ with respect to $x$: $\dfrac{dv}{dx}=-1\cdot(2Px+Q)^{-2}\cdot 2P=\dfrac{-2P}{(2Px+Q)^2}$. …

Concept and Intuition

When acceleration is proportional to velocity (and opposing it), the velocity decays exponentially with time, exactly like radioactive decay or RC discharge. There's no fixed 'half life' formula to memorize here for arbitrary factors — we derive it directly from the differential equation.

Step-by-Step Solution

1. Start with $a = \dfrac{dv}{dt} = -kv$.
2. Separate variables: $\dfrac{dv}{v} = -k\,dt$.
3. Integrate both sides from initial velocity $v_0 = v$ (at $t=0$) to $v/2$ (at time $t$): $\displaystyle\int_{v}^{v/2} \frac{dv'}{v'} = -k\int_0^t dt'$
4. LHS: $\ln(v/2) - \ln(v) = \ln(1/2) = -\ln 2$.
5. RHS: $-kt$. …

Concept and Intuition

When acceleration is not constant but a known function of time, velocity is obtained by integrating $a(t)$ with respect to time, using the initial condition to fix the constant of integration.

Step-by-Step Solution

1. Given $a=\dfrac{dv}{dt}=6t$.
2. Integrate both sides: $v(t)=\int 6t\,dt = 3t^2+C$.
3. Since the particle starts from rest, $v(0)=0 \Rightarrow C=0$.
4. So $v(t)=3t^2$.
5. At $t=2\,s$: $v=3(2)^2=3\times4=12\,ms^{-1}$.

Common Mistakes …

Concept and Intuition

A very common mix-up is to think "acceleration is decreasing, so the object must be slowing down." That confuses the rate of change of velocity with velocity itself. As long as $a(t) = \dfrac{dv}{dt} > 0$, $v$ is climbing — even if $a$ itself is falling, the velocity is still gaining, just gaining more slowly near $t_2$ than near $t_1$. The object only starts to slow down once $a$ actually crosses to negative values.

Step-by-Step Solution

1. Velocity at $t_2$ compared to $t_1$ is given by $v(t_2) = v(t_1) + \displaystyle\int_{t_1}^{t_2} a\,dt$. The integral (area under the $a$–$t$ curve) is positive as long as $a(t) \ge 0$ throughout $[t_1, t_2]$, and it is positive definite (strictly increasing $v$) as long as $a(t) > 0$ almost everywhere in that interval.
2. Graph I: straight line, positive slope, $a>0$ at $t=0$ and growing — clearly $a>0$ on all of $[t_1,t_2]$. So $v$ increases, and increases at a growing rate. …

Concept and Intuition

When a relation is given as $t$ in terms of $x$ rather than the usual $x(t)$, first invert it to get $x$ as an explicit function of $t$, then differentiate normally to get velocity as a function of time.

Step-by-Step Solution

1. $t=\sqrt{x}+3 \Rightarrow \sqrt{x}=t-3 \Rightarrow x=(t-3)^2$.
2. $v=\dfrac{dx}{dt}=2(t-3)$.
3. $v=0 \Rightarrow t=3$ s.
4. Substitute back: $x=(3-3)^2=0$ m.

Common Mistakes …

Concept and Intuition

When velocity is given as a function of position rather than time, acceleration is found via the chain rule $a=\dfrac{dv}{dt}=\dfrac{dv}{dx}\cdot\dfrac{dx}{dt} = v\dfrac{dv}{dx}$ — never simply $dv/dx$ alone.

Step-by-Step Solution

1. $v(x)=3x^2-4x$.
2. $\dfrac{dv}{dx} = 6x-4$.
3. $a = v\dfrac{dv}{dx} = (3x^2-4x)(6x-4)$.

Common Mistakes …

Concept and Intuition

Since $a=v\dfrac{dv}{dx}$, we have $v\,dv = a\,dx$, so integrating both sides: $\dfrac{v_f^2-v_i^2}{2} = \int a\,dx = \text{area under the } a\text{–}x \text{ graph}$. This lets us read the answer directly off the graph's geometry without needing $a(t)$.

Step-by-Step Solution

1. Break the a–x graph into three regions and find each area:
   • $x=0$ to $0.4$: rectangle, $a=0.4$, area $=0.4\times0.4=0.16$.
   • $x=0.4$ to $0.8$: trapezoid from $a=0.4$ down to $a=0.2$, width $0.4$: area $=\dfrac{0.4+0.2}{2}\times0.4=0.12$.
   • $x=0.8$ to $1.4$: rectangle, $a=0.2$, width $0.6$: area $=0.2\times0.6=0.12$.
2. Total area $=0.16+0.12+0.12=0.40$. …

Concept and Intuition

This is a separable first-order ODE relating velocity to position. Rearrange so all $x$-terms are on one side and $t$-terms on the other, then integrate and apply the initial condition to fix the constant.

Step-by-Step Solution

1. $v=\dfrac{dx}{dt}=e^{-\beta x}$.
2. Separate variables: $e^{\beta x}\,dx=dt$.
3. Integrate both sides: $\displaystyle\int e^{\beta x}dx=\int dt \Rightarrow \dfrac{1}{\beta}e^{\beta x}=t+C$.
4. Apply $x=0$ at $t=0$: $\dfrac1\beta e^0=0+C\Rightarrow C=\dfrac1\beta$.
5. So $\dfrac1\beta e^{\beta x}=t+\dfrac1\beta \Rightarrow e^{\beta x}=\beta t+1$. …

Concept and Intuition

The second derivative of $x$ with respect to $t$ is the acceleration: $a = d^2x/dt^2$. So the sign of acceleration directly controls the concavity (curvature) of the position-time graph, not its slope's sign. A negative acceleration means the graph curves downward (concave down) everywhere, like $x=v_0t-\tfrac12|a|t^2$, a downward parabola. A straight line has zero curvature (zero acceleration) regardless of its slope.

Step-by-Step Solution

1. For constant acceleration $a$ (here $a<0$), position is $x(t) = x_0 + v_0 t + \tfrac12 a t^2$.
2. Since $a<0$, the coefficient of $t^2$ is negative, so the curve is a downward-opening parabola — concave down everywhere.
3. Evaluate each option:
   • (A) Dome/inverted-U starting at the origin, rising to a peak, then falling: this is exactly a downward parabola through the origin — concave down. ✓
   • (B) Straight line with negative slope: $d^2x/dt^2=0$ (zero acceleration, just negative velocity). ✗ …

The key idea is simple: on a velocity-time graph, displacement equals the signed area — area above the time axis is positive (motion forward), area below is negative (motion backward). The question asks for displacement from 0 to t₄, so we only consider regions that fall within that time interval. Region E lies entirely after t₄, so it is irrelevant.

Let’s walk through it step by step.

1. Identify which regions lie between 0 and t₄.

   The time axis is marked 0, t₁, t₂, t₃, t₄, t₅.

   • Region A: from 0 to t₁ (above axis) → inside interval.
   • Region B: from t₁ to t₂ (below axis) → inside interval.
   • Region C: from t₂ to t₃ (above axis) → inside interval.
   • Region D: from t₃ to t₄ (below axis) → inside interval.
   • Region E: from t₄ to t₅ (above axis) → outside the interval 0 to t₄. So only A, B, C, D matter.

2. Assign signs to each region.

   Displacement = sum of signed areas.

   • Above axis → positive: A and C.
   • Below axis → negative: B and D. Therefore, displacement = $+A - B + C - D$.

3. Check the options.

   • (A) includes E, which is wrong.
   • (B) is exactly $A - B + C - D$ — correct.
   • (C) treats B and D as positive, which is wrong. …

Concept and Intuition

For 1-D motion with constant acceleration $a$ starting from rest ($u=0$) at $s=0$, the kinematic equation $v^2=u^2+2as$ reduces to $v^2=2as$, i.e. $s=\dfrac{v^2}{2a}$. This says $s$ is proportional to $v^2$, not to $v$ — so on an $s$ (vertical) vs. $v$ (horizontal) graph, we get a parabola opening upward through the origin, not a straight line, and it gets steeper (not flatter) as $v$ grows.

Step-by-Step Solution

1. The object starts at $s=0$ with $v=0$ and undergoes constant acceleration $a$ along its direction of motion.
2. Kinematics: $v^2=u^2+2as=2as$ (since $u=0$).
3. Solve for $s$: $s=\dfrac{v^2}{2a}$, i.e. $s$ is a quadratic (parabolic) function of $v$.
4. Since the coefficient $1/(2a)$ is positive, the parabola opens upward, passing through the origin, and its slope $ds/dv=v/a$ increases as $v$ increases — i.e., it curves upward more steeply, not flattening out. …