# Acceleration Word Problems

## Introduction to Acceleration Word Problems

The velocity of a particle may change as it (the particle) moves; then the particle is said to have been accelerated. Suppose a particle has the instantaneous velocity v1 at a time t1 and its velocity changed to v2 at time t2 . Then the change in velocity is v2 – v1 ; this change in velocity occurred in time interval t2 – t1 . Therefore the average acceleration is given by

a  =  v2 – v1 / t2 – t1  =   `Deltav`/ `Deltat` =   Change in velocity / time taken.

The acceleration is a vector . Its units in SI are ms-2 .

## Acceleration Word Problems

Problem :  A car accelerates from rest at a constant rate `alpha` for some time, after which it decelerates at a constant rate `beta` to come to rest. If the total time elapsed is 1 second, evaluate the maximum velocity reached.

Solution :  Let t1 be the time of acceleration and t2 that of deceleration for the car. The total time is

t  =  t1 + t2   …………….(1)

As the acceleration or decelerations are constant the velocity – time graph is a straight line with +ve slope and -ve slopes respectively. The slope of the line gives the acceleration `alpha` .

`:.`   `alpha`   =  slope of the line OA  =  Vmax  /  t1

`rArr`  t1  = Vmax  / `alpha`       ……………………(2)

The slope of AB gives the deceleration `beta`

`:.` `beta` = -slope of AB = Vmax / t2 =  Vmax  / `beta` .  ………………(3)

From (1) , (2)  and  (3)

t  =  t1 + t2  =  Vmax / `alpha`  +  Vmax / `beta`

t  = Vmax  [ `alpha` + `beta` /  `alpha“beta` ] .

`:.`  Vmax   =  `alpha“beta`t /  `alpha`+`beta` .

## More Acceleration Word Problems:

Problem :  A particle moving along a straight line with initial velocity u, and acceleration a continues its motion for n seconds. What is the distance covered by it in the last nth second ?

Solution :  Let the body start its motion some where else with initial velocity u and acceleration a, it is at A after (n-2) s ; at B after (n-1) s and at C after n seconds .  The body covers a distance BC in the nth second; i.e., the distance covered by it during the start and the completion of nth second. This is equal to the distance travelled by the particle in n seconds minus the distance travelled in (n-1) seconds.

BC =  Distance covered in n seconds – Distance covered in (n-1) seconds .

= [un + 1/2an2 ]  –  [ u(n-1) + 1/2a(n-1)2 ].

= un + 1/2an2  –  un + u  – 1/2an2  + an – 1/2 a

BC  = u + a[n – 1/2]

Hence the distance covered by the body in nth second

Sn = u + a[n – 1/2] .