Calculating acceleration

The English letter ‘a’ is used to denote acceleration and a _avg stands for average acceleration

When you compare the velocity change over time, you get the rate of velocity change for a given unit of time or you get the value of average acceleration for that time interval. 

So, Average acceleration = change in velocity/difference in time
                                  

                                          \(a_{avg} = {\Delta v \over \Delta t}\)
                         
                                     \(a_{avg} = { v_2 - v_1 \over t_2-t_1}\)

where, v₁ = initial velocity
             v₂ = final velocity
              t₁ = time at the starting point
              t₂ = time at the end point

∆v and ∆t are change in velocity and time respectively for the part of the journey.

The slower the rate of velocity change or acceleration, smoother is your journey; the faster the rate of velocity change or acceleration your journey will be uncomfortable.

A train is running with a speed of 60 kilometer/second. At 12:05 p.m., the driver applies the brakes to stop the train at the platform at a constant rate. At 12:10 p.m. the train stops at the platform. What should be the rate of change of the velocity (acceleration) so that train stops in 5 minutes?

We have discussed the equation of average speed. Here we will apply the same equation and calculate the rate of change of velocity i.e. acceleration. The velocity of the train reduces to 0 km/hour from 60 km/hour.

Equation of average acceleration is,        

                                                                   \(a_{avg} = {\Delta v \over \Delta t}\)

                                                                   a_avg = (0-60) kilometer/hour / 5 minute

                                                                    a_avg = -12 km/hour/minute

This answer tells you that the speed of train must have slowed down at the rate of 12 km/hour/minute to stop it in five minutes.

A negative sign indicates slowing or retardation of the train.

If the train slows down at a constant rate of 12 km/hour/minute then we can represent it as:

                                    Table 8.3: Velocity versus Time

 

Here are few questions for you to answer: