Examples on Acceleration

Some examples explaining the concept of acceleration are,

Example 1: If a truck accelerates from 6 m/s to 10 m/s in 10 s. Calculate its acceleration.

Solution:

Given,

• Initial Velocity, u = 6 m/s
• Final Velocity, v = 10 m/s
• Time taken, t = 10 s

We have to find Acceleration ‘a’

Acceleration, a = (v – u) / t

                        = (10 m/s – 6  m/s) / 10 s

                        = 0.4 m/s²

Thus, the acceleration of the truck is 0.4 m/s².

Example 2: If a ball is released from the terrace of a building to the ground. If the ball took 6 s to touch the ground. Find the height of the terrace from the ground.

Solution:

Given,

• Initial Velocity u = 0 {as the ball was at rest}
• Time taken by the ball to touch the ground t = 6 seconds 
• Acceleration due to gravity a = g = 9.8 m/s²
• Distance traveled by stone = Height of bridge  = s

Distance covered by the ball from the terrace to the ground

[Tex]s=ut+\frac{1}{2}gt^2 [/Tex]

[Tex]s = 0 + \frac{1}{2} × 9.8 × 36 = 176.4 m [/Tex]

Therefore, 

Distance of the terrace from the ground is 176.4 m.

Example 3: If a man is driving the car at 108 km/h slow down and bring it to 72 km/h in 5 s. Calculate the retardation of the car?

Solution: 

Given,

• Initial velocity, u = 108 km/h or [Tex]108\times\frac{5}{18}=30\ m/s [/Tex]
• Final velocity, v = 72 km/h or [Tex]72\times\frac{5}{18}=20\ m/s [/Tex]
• Time taken, t = 5 seconds

Therefore, acceleration is,

[Tex]\begin{aligned}a&=\dfrac{v\ -\ u}{t}\\ &=\frac{20\ -\ 30}{5}\\  &= -2\ m/s^2\end{aligned} [/Tex]

Negative sign shows retardation.

Example 4: If a car moves from rest and then accelerates uniformly at the rate of 7.5 m/s² for 10 s. Find the velocity of the train in 10 s.

Solution:

Given,

• Initial velocity u = 0 {as the car was at rest}
• Acceleration a = 7.5 m/s²
• Time t = 10 s

v = u + at

   = 0 + 7.5 × 10

   = 75 m/s

Example 5: If an object moves along the x-axis according to the relation x = 1 – 2t + 3t², where x is in meters and t is in seconds. Calculate the acceleration of the body when t = 3s.

Solution:

Given,

• x = 1 – 2t + 3t²

Velocity, v = dx/dt  

                = d/dt {1 – 2t + 3t²}

                = -2 + 6t

Therefore,

Acceleration a = dv/dt = d/dt {-2 + 6t}

a = 6 m/s² 

Acceleration is defined as the rate of change in velocity. This implies that if an object’s velocity is increasing or decreasing, then the object is accelerating. Acceleration has both magnitude and direction, therefore it is a Vector quantity. According to Newton’s Second Law of Motion, acceleration is defined as the ratio of the force applied to the object to the mass of the object. 

Let’s understand more about acceleration and related concepts like Acceleration Formula, its Unit, Types, Graphs, Solved Examples, and FAQs, in this article!