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/s^{2}
(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?