Question

A particle moves on a straight line such that product of its acceleration a and cube of velocity v is constant .The distance moved by a particle in time 't'

Answer 1

Motion in a Straight Line with Acceleration :
Velocity of a body is defined as the time rate of displacement, where as acceleration is defined as the time rate of change of velocity.Acceleration is a vector quantity. The motion may be uniformly accelerated motion or it may be non-uniformly accelerated, depending on how the velocity changes with time.

Uniform Acceleration 

The acceleration of a body is said to be uniform if its velocity changes by equal amounts in equal intervals.

Non-Uniform Acceleration

The acceleration of a body is said to be non-uniform if its velocity changes by unequal amounts in equal intervals of time.

Average velocity



Average acceleration



Illustration:

A particle moves with a velocity v(t) = (1/2)kt2 along a straight line. Find the average speed of the particle in time T.

Solution:



Illustration:

A particle having initial velocity is moving with a constant acceleration 'a' for a time t.

(a)Find the displacement of the particle in the last 1 second.

(b)Evaluate it for u = 2 m/s, a = 1 m/s2 and t = 5 sec.

Solution:

(a)  The displacement of a particle at time t is given s = ut + 1/2at2

At time (t - 1), the displacement of a particle is given by

S' = u (t-1) + 1/2a(t-1)2

 So, Displacement in the last 1 second is,

St = S - S'

= ut + 1/2 at2 – [u(t-1)+1/2 a(t-1)2 ]

= ut + 1/2at2 - ut + u - 1/2a(t - 1)2

= 1/2at2 + u - 1/2 a (t+1-2t) =  1/2at2 + u - 1/2at2 - a/2 + at

S = u + a/2(2t - 1)

(b) Putting the values of u = 2 m/s, a = 1 m/s2 and t = 5 sec, we get

S = 2 + 1/2(2 x 5 - 1) = 2 + 1/2 x 9

= 2 + 4.5 = 6.5 m   

Illustration:

Position of a particle moving along x-axis is given by x = 3t - 4t2 + t3, where x is in meters and t in seconds.

(a)Find the position of the particle at t = 2 s.

(b)Find the displacement of the particle in the time interval from t = 0 to t = 4 s.

(c)Find the average velocity of the particle in the time interval from t = 2s to t=4s.

(d)Find the velocity of the particle at t = 2 s.

Solution:

(a) x(t) = 3t - 4t2 + t3

=> x(2) = 3 x 2 - 4 x (2)2 + (2)3 = 6 - 4 x 4 + 8 = -2m.

 (b) x(o) = 0

X(4) = 3 x 4 - 4 x (4)2 + (4)3 = 12 m.

Displacement = x(4) - x(0) = 12 m.

(c) < v > = X(4)X(2)/(4-2) = (12-(-2))/2 m/s = 7 m/s

(d) dx/dt = 3 - 8t + 3t2

v(2) (dx/dt)2 = 3 - 8 x 2 + 3 x (2)2 = -1m/s

Illustration:

Two trains take 3 sec to pass one another when going in the opposite direction but only 2.5 sec if the speed of the one is increased by 50%. The time one would take to pass the other when going in the same direction at their original speed is

(a) 10 sec                 (b) 12 sec

(c) 15 sec                 (d) 18 sec

Solution:

Using the equation,

t = d/vr

We have,

3 = d/v1+v2

2.5 = d/1.5v1+v2

Solving we get,

v1 = 2d/15 and v2 = d/5

When they are going in same direction,

vr = v2 – v1 = d/15

Thus, t = d/vr = d/(d/15) = 15 s

From the above observation we conclude that, option (c) is correct. 

Analysis of Uniformly Accelerated Motion



Case-I:
 For uniformly accelerated motion with initial velocity u and initial position x0.

Velocity Time Graph

              

In every case tanθ = a0

Position Time Graph

       

Initial position x of the body in every case is x0 (> 0)

Case II:

For uniformly retarded motion with initial velocity u and initial position x0.                                         

Velocity Time Graph

                  

In every case tanθ  = -a0

Position Time Graph

             

Initial position x of the body in every case is x0 (> 0)

Illustration:



A particle is moving rectilinearly with a time varying acceleration a = 4 - 2t, where a is in m/s2 and t is in sec. If the particle is starting its motion with a velocity of -3 m/s from x = 0. Draw a-t, v-t and x-t curve for the particle.

Solution:

a = 4-2t



v = 4t-t2-3



x = 2t2 – t3/3 – 3t 

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