Learn graphs of various parameters of motion & difference between accelerating and non-accelerating motion. Practice equation to calculate of acceleration, displacement from v-t graph and conversion from v-t to s-t graph.
For example :
The path of the car can be shown as follows -
The graph between position and time can be represented as -
A Particle moves in negative y-direction.
B Particle moves in positive y-direction.
C Particle first moves in positive y-direction and then moves in negative y- direction.
D Particle first moves in negative y-direction and then moves in positive y-direction.
A Non-accelerating motion \(\rightarrow\) Accelerating motion \(\rightarrow\) Accelerating motion
B Accelerating motion \(\rightarrow\) Non-accelerating motion \(\rightarrow\) Accelerating motion
C Accelerating motion \(\rightarrow\) Accelerating motion \(\rightarrow\) Non-accelerating motion
D Non accelerating motion \(\rightarrow\) Accelerating motion \(\rightarrow\) Non-accelerating motion
v_{1} = slope of graph (1) = \(tan\,\theta_1\)
v_{2} = slope of graph (2) = \(tan\,\theta_2\)
\(v_1>v_2\)
\(\Delta t_1<\Delta t_2\)
A Velocity of particle (1) is more than velocity of particle (2).
B To reach at certain point, particle (2) will take more time.
C To reach at certain point, particle (1) will take more time.
D Both the particles are traveling along positive directions.
Here, \(\theta_1>\theta_2\)
\(\therefore\,tan\theta_1>tan\theta_2\)
\(a_1>a_2\)
\(\Delta t_1\,<\,\Delta t_2\)
It means particle (2) takes more time.
Displacement = Area under v-t graph
A Both will have same displacement in equal time
B Acceleration of particle (2) is less than acceleration of particle (1)
C Both are having equal accelerations
D Acceleration of particle (2) is more than acceleration of particle (1)
Acceleration = \(\dfrac {\Delta v} {\Delta t} \)= \(\dfrac { v_3–v_1} {t_2–0} \)
Now, from the graph,
\(\dfrac {\Delta v} {\Delta t} \)= tan\(\theta\) \(\implies\) slope
\(\therefore\,\text{Acceleration = slope of v-t graph}\)
Points to remember :
\(tan\theta\) is positive when \(0^0<\theta< 90^0\)
\(tan\theta\) is negative when \(90^0<\theta< 180^0\)
\(tan\theta\) is \(\infty\) when \(\theta=90^0\)
2. If \(\theta\) is obtuse, the slope will be negative.
3. If \(\theta\) is \(90^0\), the slope will be infinite.
A \(3\,m/s^2\)
B \(4\,m/s^2\)
C \(2\,m/s^2\)
D \(5\,m/s^2\)
Displacement in the time interval \(\Delta t\) = Area under the v-t curve
For \(\Delta t_1\) time :
For \(\Delta t_2\) time interval :
For \(\Delta t_3\) time interval :
Velocity = 0,
\(\therefore\) Slope of s-t graph = 0
In \(\Delta t_1\) time interval
Acceleration, a > 0
slope of v-t curve > 0
Velocity will increase.
For \(\Delta t_2\) time interval
Acceleration < 0
slope of v-t graph < 0
Velocity of a particle will decrease and it will increase in negative direction.
For \(\Delta t_3\) time interval
Acceleration = 0
slope of v-t graph = 0
\(\therefore\) Velocity in \(\Delta t_3\) time interval is constant.
\(v=\dfrac{ds}{dt}\)= slope of s-t graph
\(v=\dfrac{ds}{dt}=tan\,\theta\)
a_{1} = slope of v-t graph = \(tan\,\theta_1\)
a_{2} = slope of v-t graph \(=tan\theta_2\)
a_{3} = slope of v-t graph \(=tan\theta\)
So, the a-t graph of the particle will be