Learn velocity and acceleration formula in calculus, practice for finding the acceleration & speeding up and slowing down interval from velocity graph.

Acceleration is given by **\(a(t)=\dfrac {dv}{dt}=\dfrac {d^2s}{dt^2}\)**

- A particle speeds up when velocity and acceleration both are of same sign i.e. either both are positive or both are negative.
- Speeding up \(\Rightarrow\) a > 0 or v > 0 or a < 0 and v < 0

A t > 7 and 2 < t < 4.5

B t > 9

C t > 8 or 3 < t < 5

D t < 3

- A particle slows down when velocity and acceleration both are of opposite signs i.e. a > 0 and v < 0 or a < 0 and v > 0.
- Slowing down \(\Rightarrow\) a > 0 and v < 0 or a < 0 & v > 0

A 9/2 < t < 7 and t < 2

B t > 9

C 2 < t < 5

D 3/2 < t < 5 or t < 1

- A particle is said to speed up under two conditions:

- If velocity and acceleration are both positive.
- If velocity and acceleration are both negative.

i.e., If a > 0 & v > 0 **or** a < 0 & v < 0

where, a = acceleration and v = velocity

- If velocity - time graph is given then time of speeding up is the part of graph when it is above x-axis and increasing or below x axis and decreasing.

A t > 4 and 5 < t < 6

B 5 < t < 7

C 2 < t < 5 and 0 < t < 1

D 0 < t < 6

- A particle is said to be slowing down under two conditions:

- If velocity is positive and acceleration is negative.
- If velocity is negative and acceleration is positive.

i.e., If a > 0 & v < 0** or** a < 0 & v > 0

where, a = acceleration and v = velocity

- If velocity - time graph is given then time of slowing down is the part of graph when it is above the x-axis and decreasing or below x axis and increasing.

A 2 < t < 4 or 6 < t < 7

B t > 5

C 1 < t < 2 or 5 < t < 7

D t < 4

- When a particle is thrown vertically upwards if reaches a particular height and then comes down due to the effect of gravity,
- To find the maximum height reached by particle first fid the time at which the velocity is 0 i.e. particle is momentarily at rest then find height by putting the value.