Acceleration and its graph Questions in English

(D) The average acceleration $a_{av}$ is defined as the total change in velocity divided by the total time interval.
By definition,the change in velocity $\Delta v$ over a time interval from $t_1$ to $t_2$ is given by the integral of acceleration with respect to time: $\Delta v = \int_{t_1}^{t_2} a dt$.
The time interval is $\Delta t = t_2 - t_1$.
Therefore,the average acceleration is $a_{av} = \frac{\Delta v}{\Delta t} = \frac{\int_{t_1}^{t_2} a dt}{t_2 - t_1}$.

(B) The change in velocity $(\Delta v)$ is equal to the area under the acceleration-time $(a-t)$ graph.
From the graph, the area from $t=0$ to $t=4 \, s$ is a triangle with base $4 \, s$ and height $4 \, m/s^2$:
Area$_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 4 = 8 \, m/s$.
The area from $t=4 \, s$ to $t=6 \, s$ is a triangle below the time axis with base $(6-4) = 2 \, s$ and height $-4 \, m/s^2$:
Area$_2 = \frac{1}{2} \times 2 \times (-4) = -4 \, m/s$.
The total change in velocity is the sum of these areas:
$\Delta v = \text{Area}_1 + \text{Area}_2 = 8 + (-4) = 4 \, m/s$.
Thus, the correct option is $(b)$.

(A) The distance travelled is equal to the area under the velocity-time $(v-t)$ graph.
First, we construct the $v-t$ graph from the given acceleration-time $(a-t)$ graph:
$1$. For $0 \le t \le 1 \, s$, $a = 2 \, m/s^2$. Since $v = \int a \, dt$, $v = 2t$. At $t = 1 \, s$, $v = 2 \, m/s$.
$2$. For $1 \le t \le 2 \, s$, $a = 0$, so velocity remains constant at $v = 2 \, m/s$.
$3$. For $2 \le t \le 3 \, s$, $a = -2 \, m/s^2$. Velocity decreases from $2 \, m/s$ to $0 \, m/s$ at $t = 3 \, s$.
The resulting $v-t$ graph is a trapezoid with parallel sides of length $1 \, s$ (from $t=1$ to $t=2$) and $3 \, s$ (from $t=0$ to $t=3$), and height $2 \, m/s$.
Distance = Area of the trapezoid = $\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$
Distance = $\frac{1}{2} \times (1 + 3) \times 2 = 4 \, m$.

(A) From the given acceleration $(a)$-time $(t)$ graph,it is evident that the acceleration is positive and decreasing linearly with time.
Let the acceleration be represented as $a = a_0 - kt$,where $a_0$ is the initial acceleration and $k$ is a positive constant.
Since the particle starts from rest,the initial velocity $v(0) = 0$.
We know that $a = \frac{dv}{dt}$,so $\frac{dv}{dt} = a_0 - kt$.
Integrating both sides with respect to time $t$:
$\int_{0}^{v} dv = \int_{0}^{t} (a_0 - kt) dt$
$v = a_0 t - \frac{1}{2} kt^2$.
This equation represents a downward-opening parabola starting from the origin $(0,0)$.
As time $t$ increases,the slope of the velocity-time graph,which is the acceleration $a$,decreases. Therefore,the velocity-time graph should be a curve that starts with a positive slope and becomes flatter (slope decreases) as time approaches $T$. This corresponds to the graph shown in option $(A)$.

(A) The change in velocity of a particle is equal to the area under the acceleration-time $(a-t)$ graph.
Given:
Initial velocity, $u = -5 \, m/s$
From the graph:
$1$. The area under the $a-t$ graph from $t = 0$ to $t = 6 \, s$ is a triangle with base $6 \, s$ and height $10 \, m/s^2$.
Area$_1 = \frac{1}{2} \times 6 \times 10 = 30 \, m/s$.
$2$. The area under the $a-t$ graph from $t = 6 \, s$ to $t = 8 \, s$ is a triangle with base $(8 - 6) = 2 \, s$ and height $-10 \, m/s^2$.
Area$_2 = \frac{1}{2} \times 2 \times (-10) = -10 \, m/s$.
Net area under the graph = Area$_1 +$ Area$_2 = 30 + (-10) = 20 \, m/s$.
Since $\Delta v = v - u = \text{Net Area}$, we have:
$v - (-5) = 20$
$v + 5 = 20$
$v = 15 \, m/s$.
Therefore, the velocity at $t = 8 \, s$ is $15 \, m/s$.

(C) Given the velocity equation: $v = \frac{t^2}{10} + 20$.
To find the acceleration $a$,we differentiate the velocity with respect to time $t$:
$a = \frac{dv}{dt} = \frac{d}{dt} \left( \frac{t^2}{10} + 20 \right)$.
Applying the power rule:
$a = \frac{2t}{10} + 0 = \frac{t}{5}$.
Since the acceleration $a$ depends on time $t$ $(a \propto t)$,it is not constant. Therefore,the body is undergoing non-uniform acceleration.

(B) The position of the body is given by the equation $x = 4t^2 - 12t$.
To find the velocity $v$,we differentiate the position $x$ with respect to time $t$:
$v = \frac{dx}{dt} = \frac{d}{dt}(4t^2 - 12t) = 8t - 12 \, m/s$.
To find the acceleration $a$,we differentiate the velocity $v$ with respect to time $t$:
$a = \frac{dv}{dt} = \frac{d}{dt}(8t - 12) = 8 \, m/s^2$.
Therefore,the acceleration of the particle is $8 \, m/s^2$.

(B) Given position function: $x = -3t^3 + 18t^2 + 16t$.
Velocity $v$ is the first derivative of position with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(-3t^3 + 18t^2 + 16t) = -9t^2 + 36t + 16$.
Acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(-9t^2 + 36t + 16) = -18t + 36$.
Set acceleration to zero to find the time: $-18t + 36 = 0 \implies 18t = 36 \implies t = 2 \,s$.
Substitute $t = 2 \,s$ into the velocity equation: $v = -9(2)^2 + 36(2) + 16$.
$v = -9(4) + 72 + 16 = -36 + 72 + 16 = 52 \,m/s$.

(B) The acceleration $(a)$ is defined as the rate of change of velocity,which corresponds to the slope of the velocity-time $(v-t)$ graph,i.e.,$a = \frac{dv}{dt}$.
$1$. In the first interval,the velocity increases linearly with time. Since the slope is constant and positive,the acceleration is constant and positive.
$2$. In the second interval,the velocity is constant. Since the slope of a horizontal line is zero,the acceleration is zero.
$3$. In the third interval,the velocity decreases linearly with time. Since the slope is constant and negative,the acceleration is constant and negative.
Comparing this with the given options,the graph that shows a positive constant acceleration,followed by zero acceleration,and then a negative constant acceleration is represented by option $B$.

(D) Given the displacement equation: $x = c_0(t^2 - 2) + c(t - 2)^2$.
To find the velocity $v$,we differentiate $x$ with respect to $t$:
$v = \frac{dx}{dt} = \frac{d}{dt}[c_0(t^2 - 2) + c(t - 2)^2] = 2c_0t + 2c(t - 2)$.
To find the acceleration $a$,we differentiate $v$ with respect to $t$:
$a = \frac{dv}{dt} = \frac{d}{dt}[2c_0t + 2c(t - 2)] = 2c_0 + 2c = 2(c_0 + c)$.
Thus,the acceleration of the particle is $2(c + c_0)$.

(C) Let the constant acceleration be $a = k$,where $k$ is a positive constant.
We know that acceleration $a = v \frac{dv}{ds}$.
Substituting $a = k$,we get $k = v \frac{dv}{ds}$.
Rearranging the terms,we have $k \, ds = v \, dv$.
Integrating both sides with initial conditions $s=0$ at $v=0$ (assuming the particle starts from rest):
$\int_0^{s} k \, ds = \int_0^{v} v \, dv$
$ks = \frac{v^2}{2}$
$v^2 = 2ks$
This equation represents a parabola opening along the $s$-axis. Since $s = \frac{v^2}{2k}$,the graph of $s$ versus $v$ is a parabola symmetric about the $s$-axis,opening towards the positive $s$-direction. Looking at the provided options,Graph $C$ represents this relationship where $s$ increases quadratically with $v$.

(A) The change in velocity $(\Delta v)$ is equal to the area under the acceleration-time $(a-t)$ graph.
Given that the body starts from rest, the initial velocity $(u = 0 \ m/s)$.
The area under the $a-t$ graph is a right-angled triangle with base $(b = 10 \ s)$ and height $(h = 8 \ m/s^2)$.
Area = $\frac{1}{2} \times \text{base} \times \text{height}$
Area = $\frac{1}{2} \times 10 \ s \times 8 \ m/s^2 = 40 \ m/s$.
Since $\Delta v = v_{max} - u = 40 \ m/s$ and $u = 0 \ m/s$, the maximum speed is $v_{max} = 40 \ m/s$.

(C) Given the position function: $x = at^2 - bt^3$
Velocity $v$ is the first derivative of position with respect to time: $v = \frac{dx}{dt} = 2at - 3bt^2$
Acceleration $a_{acc}$ is the derivative of velocity with respect to time: $a_{acc} = \frac{dv}{dt} = 2a - 6bt$
To find the time when acceleration is zero,set $a_{acc} = 0$:
$0 = 2a - 6bt$
$6bt = 2a$
$t = \frac{2a}{6b} = \frac{a}{3b}$

(B) The change in velocity of a particle is equal to the area under the acceleration-time $(a-t)$ graph.
Given that the particle starts from rest,the initial velocity $u = 0$.
The area under the $a-t$ graph is a triangle with base $b = 10 \ s$ and height $h = 8 \ m/s^2$.
Area $= \frac{1}{2} \times \text{base} \times \text{height}$
Area $= \frac{1}{2} \times 10 \times 8 = 40 \ m/s$.
Since $\Delta v = v - u = \text{Area}$,and $u = 0$,we have $v = 40 \ m/s$.
Thus,the maximum speed of the particle is $40 \ m/s$.

(A) The equation of motion for a particle with constant acceleration is given by $x(t) = x_0 + v_0 t + \frac{1}{2} a t^2$.
If the acceleration $a$ is negative,the graph of $x$ versus $t$ is a parabola opening downwards.
In a position-time graph,the slope represents the velocity $(v = \frac{dx}{dt})$.
For a downward-opening parabola,the slope starts as positive,decreases to zero at the peak (where the particle momentarily stops),and then becomes negative as the particle moves in the opposite direction.
This behavior corresponds to a constant negative acceleration.
Therefore,the graph that represents a downward-opening parabola is the correct one,which is shown in option $(A)$.

(C) The displacement of the particle is given by $s = t^3 - 6t^2 + 18t + 9$.
The velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{ds}{dt} = 3t^2 - 12t + 18$.
To find the minimum velocity,we find the derivative of velocity with respect to time and set it to zero: $\frac{dv}{dt} = 6t - 12$.
Setting $\frac{dv}{dt} = 0$,we get $6t - 12 = 0$,which implies $t = 2 \ s$.
Now,substitute $t = 2 \ s$ into the velocity equation to find the minimum velocity:
$v_{min} = 3(2)^2 - 12(2) + 18 = 3(4) - 24 + 18 = 12 - 24 + 18 = 6 \ m \ s^{-1}$.

(B) We know that $a = v \frac{dv}{dx}$, which implies $v dv = a dx$.
Integrating both sides, we get $\int_{u}^{v} v dv = \int_{x_1}^{x_2} a dx$.
$\frac{v^2 - u^2}{2} = \text{Area under the } a-x \text{ graph}$.
$v^2 = u^2 + 2 \times (\text{Area under the } a-x \text{ graph})$.
The area under the $a-x$ graph from $x=0$ to $x=1.4$ consists of a rectangle, a trapezoid, and another rectangle:
Area $1$ (from $x=0$ to $0.4$): $0.4 \times 0.4 = 0.16$.
Area $2$ (from $x=0.4$ to $0.8$): $\frac{1}{2} \times (0.4 + 0.2) \times 0.4 = 0.12$.
Area $3$ (from $x=0.8$ to $1.4$): $0.2 \times 0.6 = 0.12$.
Total Area $= 0.16 + 0.12 + 0.12 = 0.4$.
Given $u = 0.8 \,ms^{-1}$, so $u^2 = 0.64$.
$v^2 = 0.64 + 2 \times (0.4) = 0.64 + 0.8 = 1.44$.
$v = \sqrt{1.44} = 1.2 \,ms^{-1}$.

(D) Given, velocity $v = 2t^2 - 8t + 15$.
We know that instantaneous acceleration $a$ is the derivative of velocity with respect to time $t$, i.e., $a = \frac{dv}{dt}$.
Differentiating the given expression for velocity with respect to $t$:
$a = \frac{d}{dt}(2t^2 - 8t + 15) = 4t - 8$.
Now, substitute the value $t = 5 \,s$ into the expression for acceleration:
$a = 4(5) - 8 = 20 - 8 = 12 \,ms^{-2}$.
Therefore, the instantaneous acceleration at $t = 5 \,s$ is $12 \,ms^{-2}$.

(A) Given the displacement equation: $S = 4t^3 - 8t^2 + 5t + 4$.
Velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{dS}{dt} = \frac{d}{dt}(4t^3 - 8t^2 + 5t + 4) = 12t^2 - 16t + 5$.
Acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(12t^2 - 16t + 5) = 24t - 16$.
To find the acceleration at $t = 2 \ s$,substitute $t = 2$ into the acceleration equation: $a = 24(2) - 16 = 48 - 16 = 32 \ m \ s^{-2}$.

(A) Given the displacement equation: $x = 2t^2 + t + 5$.
Velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(2t^2 + t + 5) = 4t + 1$.
Acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(4t + 1) = 4 \ m \cdot s^{-2}$.
Since the acceleration is constant,the acceleration at $t = 2 \ s$ is $4 \ m \cdot s^{-2}$.

(A) The position of the particle is given by $s(t) = 3t^3 + 7t^2 + 3t + 8$.
Velocity $v(t)$ is the first derivative of position with respect to time: $v(t) = \frac{ds}{dt} = \frac{d}{dt}(3t^3 + 7t^2 + 3t + 8) = 9t^2 + 14t + 3$.
Acceleration $a(t)$ is the derivative of velocity with respect to time: $a(t) = \frac{dv}{dt} = \frac{d}{dt}(9t^2 + 14t + 3) = 18t + 14$.
At $t = 1 \ s$,the acceleration is $a(1) = 18(1) + 14 = 18 + 14 = 32 \ m/s^2$.

(C) The velocity of the particle is given by $v = 3x^2 - 4x$.
Acceleration $a$ is defined as the rate of change of velocity with respect to time,which can be expressed in terms of position $x$ as $a = v \cdot \frac{dv}{dx}$.
First,differentiate $v$ with respect to $x$: $\frac{dv}{dx} = \frac{d}{dx}(3x^2 - 4x) = 6x - 4$.
Now,substitute $v$ and $\frac{dv}{dx}$ into the acceleration formula: $a = (3x^2 - 4x)(6x - 4)$.
Thus,the correct expression for acceleration is $(3x^2 - 4x)(6x - 4)$.

(C) Given the position function: $x = (2t - 3)^2$.
Expanding the expression: $x = 4t^2 - 12t + 9$.
Velocity $v$ is the first derivative of position with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(4t^2 - 12t + 9) = 8t - 12$.
Acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(8t - 12) = 8 \,m/s^2$.
Since the acceleration is constant,at $t = 2 \,s$,the acceleration remains $8 \,m/s^2$.

(B) The particle starts from rest,so its initial velocity $u = 0 \ m \ s^{-1}$.
In an acceleration-time $(a-t)$ graph,the change in velocity $(\Delta v)$ is equal to the area under the curve.
Since the acceleration is positive throughout the interval from $t = 0 \ s$ to $t = 15 \ s$,the velocity of the particle increases continuously.
Therefore,the maximum speed is attained at $t = 15 \ s$.
The area under the $a-t$ graph is the area of the right-angled triangle with base $15 \ s$ and height $10 \ m \ s^{-2}$.
$\Delta v = \text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$
$\Delta v = \frac{1}{2} \times 15 \ s \times 10 \ m \ s^{-2} = 75 \ m \ s^{-1}$.
Since $v_{max} = u + \Delta v = 0 + 75 \ m \ s^{-1} = 75 \ m \ s^{-1}$.

(B) Given the relation: $t = 2x^2 + 3x$.
To find velocity $v$,differentiate $t$ with respect to $x$: $\frac{dt}{dx} = 4x + 3$.
Since $v = \frac{dx}{dt}$,we have $v = \frac{1}{4x + 3} = (4x + 3)^{-1}$.
To find acceleration $a$,differentiate $v$ with respect to $t$: $a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{dv}{dx} \cdot v$.
$\frac{dv}{dx} = -1(4x + 3)^{-2} \cdot 4 = -\frac{4}{(4x + 3)^2}$.
Thus,$a = -\frac{4}{(4x + 3)^2} \cdot \frac{1}{4x + 3} = -\frac{4}{(4x + 3)^3}$.
Given displacement $x = 25 \ cm = 0.25 \ m = \frac{1}{4} \ m$.
Substitute $x = \frac{1}{4}$ into the acceleration formula: $a = -\frac{4}{(4(1/4) + 3)^3} = -\frac{4}{(1 + 3)^3} = -\frac{4}{4^3} = -\frac{4}{64} = -\frac{1}{16} \ ms^{-2}$.

(B) The acceleration vector is given by the material derivative of velocity: $\vec{a} = \frac{d\vec{v}}{dt} = (\vec{v} \cdot \nabla) \vec{v}$.
Given $\vec{v} = -x\hat{i} + 2y\hat{j} - z\hat{k}$,we calculate the components:
$a_x = v_x \frac{\partial v_x}{\partial x} + v_y \frac{\partial v_x}{\partial y} + v_z \frac{\partial v_x}{\partial z} = (-x)(-1) + (2y)(0) + (-z)(0) = x$.
$a_y = v_x \frac{\partial v_y}{\partial x} + v_y \frac{\partial v_y}{\partial y} + v_z \frac{\partial v_y}{\partial z} = (-x)(0) + (2y)(2) + (-z)(0) = 4y$.
$a_z = v_x \frac{\partial v_z}{\partial x} + v_y \frac{\partial v_z}{\partial y} + v_z \frac{\partial v_z}{\partial z} = (-x)(0) + (2y)(0) + (-z)(-1) = z$.
Thus,$\vec{a} = x\hat{i} + 4y\hat{j} + z\hat{k}$.
At point $(1, 2, 4)$,$\vec{a} = 1\hat{i} + 4(2)\hat{j} + 4\hat{k} = 1\hat{i} + 8\hat{j} + 4\hat{k}$.
The magnitude is $|\vec{a}| = \sqrt{1^2 + 8^2 + 4^2} = \sqrt{1 + 64 + 16} = \sqrt{81} = 9 \text{ m/s}^2$.