## 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 v_{1} at a time t_{1} and its velocity changed to v_{2} at time t_{2} . Then the change in velocity is v_{2} – v_{1} ; this change in velocity occurred in time interval t_{2} – t_{1} . Therefore the average acceleration is given by

a = v_{2} – v_{1 / }t_{2} – t_{1 = } `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 t_{1} be the time of acceleration and t_{2} that of deceleration for the car. The total time is

t = t_{1} + t_{2} …………….(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 = V_{max }/ t_{1}

`rArr` t_{1} = V_{max }/ `alpha` ……………………(2)

The slope of AB gives the deceleration `beta`

`:.` `beta` = -slope of AB = V_{max} / t_{2} = V_{max }/ `beta` . ………………(3)

From (1) , (2) and (3)

t = t_{1} + t_{2} = V_{max} / `alpha` + V_{max }/ `beta`

t = V_{max }[ `alpha` + `beta` / `alpha“beta` ] .

`:.` V_{max } = `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 n^{th} 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 + 1/2an^{2} ] – [ u(n-1) + 1/2a(n-1)^{2} ].

= un + 1/2an^{2} – un + u – 1/2an^{2} + an – 1/2 a

BC = u + a[n – 1/2]

Hence the distance covered by the body in nth second

S_{n} = u + a[n – 1/2] .

can you solve this question-

find the initial velocity and acceleration of a body, if the distance traveled by that body in nth second is (5.96+0.08n)m.

please help me to solve this question as soon as possible