Practice equation to calculate of maximum height attained by a ball and velocity and displacement of a particle moving along a straight line with constant acceleration. Learn uniform and constant acceleration formula.

- Acceleration is defined as the rate of change of velocity with respect to time.

\(\vec a_{avg}=\dfrac{\Delta \vec v}{\Delta \,t}\)

- From above relation, it is clear that acceleration is also a vector quantity.

**Note : **

- Direction of acceleration vector is same as that of change in velocity vector.
- There are two ways in which velocity can change.

(a) Change in magnitude

(b) Change in direction

- So, two types of accelerations can be produced.

- In one-dimensional motion, acceleration can have only two directions, either positive or negative.
- To understand one-dimensional motion with constant acceleration, consider a particle is moving along a straight line with uniform (constant) acceleration \('a'\). If the velocity of the particle at \(t=t_1\) is \(v_0\) and at \(t=t_2\) is \(v\), then

\(v=v_0+a(t_2-t_1)\)

\(v=v_0+a(\Delta\,t)\)

- It is clear from the above equation that, if

(a) Acceleration and velocity are in same direction, magnitude of velocity ,i.e., speed of the particle will increase.

(b) Acceleration and velocity are in opposite direction, magnitude of velocity ,i.e., speed of the particle will decrease.

A velocity is \(3\,m/sec\) and acceleration is \(-5\,m/sec^2\), speed is increasing

B velocity is \(3\,m/sec\) and acceleration is \(5\,m/sec^2\), speed is increasing

C velocity is \(-3\,m/sec\) and acceleration is \(5\,m/sec^2\), speed is increasing

D velocity is \(3\,m/sec\) and acceleration is \(5\,m/sec^2\) , speed is decreasing

- Let us consider a particle, moving along a straight line with constant acceleration \('a'\).
- If the initial velocity of the particle at \(t=t_0\) is \(v_0\) , then the velocity at time \(t\) will be

\(a=\dfrac{v-v_0}{t-t_0}\)

\(\Rightarrow\;v-v_0=a(t-t_0)\)

\(\Rightarrow\;v=v_0+a(t-t_0)\)

\(\Rightarrow\;v=v_0+a(\Delta t)\)

- Since, velocity of the particle is varying uniformly, we can say

\(v_{avg}=\dfrac{v+v_0}{2}\)

- Average velocity \((v_{avg})\) is given as

\(v_{avg}=\dfrac{\text{Displacement}}{\text{Time interval}}=\dfrac{\Delta x}{\Delta t}\)

\(\Rightarrow\;\Delta x=v_{avg}(\Delta t)\)

\(\Rightarrow\;\Delta x=\left(\dfrac{v+v_0}{2}\right)\,\Delta t\)

Using \(v=v_0+a(\Delta t)\),

\(\Rightarrow\;\Delta x=v_0(\Delta t)+\dfrac{1}{2}a(\Delta t)^2\)

A \(61\,m/sec,\,232\,m\)

B \(6\,m/sec,\,13\,m\)

C \(68\,m/sec,\,120\,m\)

D \(50\,m/sec,\,60\,m\)

A \(5\,m/s\;,\,2\,m/s^2\)

B \(3\,m/s\;,\,8\,m/s^2\)

C \(2\,m/s\;,\,6\,m/s^2\)

D \(6\,m/s\;,\,2\,m/s^2\)

- Let us consider a particle, moving along a straight line with constant acceleration \('a'\). If the initial velocity of the particle at \(t=t_0\) is \(v_0\), then the velocity at time \(t\) will be

\(a=\dfrac{v-v_0}{t-t_0}\)

\(\Rightarrow\;v-v_0=a(t-t_0)\)

\(\Rightarrow\;v=v_0+a(t-t_0)\)

\(\Rightarrow\;v=v_0+a(\Delta t)\)

- Since, velocity of the particle is varying uniformly, we can say

\(v_{avg}=\dfrac{v+v_0}{2}\)

- Average velocity \((v_{avg})\) is given as

\(v_{avg}=\dfrac{\text{Displacement}}{\text{Time interval}}=\dfrac{\Delta x}{\Delta t}\)

\(\Rightarrow\;\Delta x=v_{avg}(\Delta t)\)

\(\Rightarrow\;\Delta x=\left(\dfrac{v+v_0}{2}\right)\,\Delta t\)

Using \(v=v_0+a(\Delta t)\)

\(\Rightarrow\;\Delta x=v_0(\Delta t)+\dfrac{1}{2}a(\Delta t)^2\)

where, \(\Delta x=x_2-x_1\)

A \(16\ m\)

B \(25\ m\)

C \(20\ m\)

D \(38\ m\)

- Consider a particle with uniform acceleration \('a\,'\) is moving along a straight line.
- If the velocity of the particle changes from \(v_0\) to \(v\) in a time interval of \(\Delta t\), the displacement in this time interval can be obtained by,

\(\Delta x=v_o(\Delta t)+\dfrac{1}{2}a(\Delta t)^2\) .....(1)

where \(\Delta t=\dfrac{v-v_0}{a}\) .....(2)

- After eliminating \(\Delta t\) from equation (1), we get

\(v^2=v_0^2+2a(\Delta x)\)

\(\Delta x=\dfrac{v^2-v_0^2}{2\,a}\)

where , \(v_0=\) Initial velocity

\(v=\) Final velocity

\(a=\) Acceleration

\(\Delta x=\) Displacement

- Various parameters of a particle projected vertically, comes under the domain of one dimensional motion as velocity of the particle is always along a straight line and its acceleration is along the line of motion.
- So, all the formulae of one-dimensional motion, are applicable for calculating the parameters.

- Assuming, the motion in upward direction is positive and the motion in downward direction is negative.
- Let, maximum height attained by the particle when projected vertically upward be \(H.\)
- At height \(H\), the velocity will be zero, i.e. \(v=0\).
- Let the initial velocity of the particle is \(v_0\) and the acceleration will be \(-g\).
- So using

\(v^2=v_o^2+2a\Delta x\) { here, v_{0 = initial velocity , v = final velocity} }

\(0=v_o^2-2gH\)

\(H_{max}=\dfrac{v_o^2}{2g}\)

A \(6\,m\)

B \(8\,m\)

C \(5\,m\)

D \(9\,m\)

A \(110\,m\)

B \(117\,m\)

C \(120\,m\)

D \(130\,m\)