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] .