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Motion in straight line with variable acceleration with Problem and solutions



Motion in straight line with variable acceleration

  • A particle may move in a straight line with a non-uniform accleration i,e variable acceleration
  • There is no direct formula for motion in it as the equation would vary in each case
  • we need to apply the basic formula to get the velocity ,acceleration in it

Formula to be used

(1) Instantanous velocity is given by
$v= \frac {dx}{dt}$
(2) Instantanous acceleration is given by
$a= \frac {dv}{dt}= v\frac {dv}{dx}$
(3) Average velocity is given by
$v_{avg} = \frac {\int_{0}^{t }vdt}{\int_{0}^{t }dt}$
(4) Average acceleration is given by
$a_{avg} = \frac {\int_{0}^{t }adt}{\int_{0}^{t }dt}$

Solved Examples

Question 1
Motion of a particle is given by the equation
$x=3t^3 + 7t^2 + 14t + 8$
where t in sec and x is in meter.
(a) Find the position of the particle at 1 sec
(b)Find the displacement of the particle in the time interval from t=0 to t=4 sec
(c) Find the value of the velocity at 2 sec
(d) Find the value of the acceleration of the particle at t=1 sec
(e) Find the average velocity of the particle in the interval t=2 sec to t=4 sec
Solution
(a) position of the particle at 1 sec is given by
$x_{1}= 3(1)^3 + 7(1)^2 + 14(1) + 8=32$ m
(b)Position of the particle at 4 sec is given by
$x_4=3(4)^3+ 7(4)^2 + 14(4) + 8=192 + 112+ 28+8=368$ m
We already know the position of the particle at t=0
Displacement between t=0 to t=4 is given by
$=x_4 - x_0=368 -32=336$m
(c) Velocity is given by
$v=\frac {dx}{dt} = 9t^2 + 14t + 14$
Now velocity at 2 sec
$v_2=9(2)^2 + 14(2) + 14=78$m/s
(d) Accleration is given by

$a=\frac {dv}{dt}=18t + 14$
Now acceleration of the particle at t=1 sec
$a_1=18(1) + 14=32$m/s2
(e)Average velocity will be give as
$=\frac {x_4 -x_2}{4-2}$
Now
$x_2=3(2)^3+ 7(2)^2 + 14(2) + 8=24+28+ 28+8=88$ m
So Average velocity
$= \frac {336-88}{2}=124$ m/s

Question 2
The distance x of a particle moving in one-direction, under the action of a constant force is related to time by equation
$t= \sqrt {x} +3$
where x is in metres, t is in seconds.
Find the displacement of the particle when its velocity is zero.
Solution
The equation of time is given as,
$t= \sqrt {x} +3$
it can be rearranged as
$\sqrt {x} =t−3$
Squarring both the sides
$x=t^2 -6t+9$
The velocity is given as,
$v=\frac {dx}{dt} =2t−6$
When the velocity is zero, it can be written as,
0=2t−6
t=3
Therefore The displacement at 3s is given as,
$\sqrt x=3-3$
$x=0$

Problems

Question 1
A particle’s position on the x axis is given by
$x = 4 - 27t + t^3$
with x in meters and t in seconds.
Find the particle’s velocity function v(t) and acceleration function a(t).

Question 2
A particle initialy moving with velocity u is subjected to a retarding force as a result, it deccelerates with $a=-k \sqrt {v}$ where v is instantanous velocity and k is positive constant.
Find the time taken by the particle to come to rest?

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