PLEASE GIVE THE SOLUTIONS WITHIN 2 HOURS TOMORROW IS MY SUBMISSION DATEQ4. A particle moves along a straight line with a constant acceleration a = +0.5 m/s2. Att= 0, it is at x = 0, and its velocity v = 0. Plot the velocity-time and position-timegraphs for the period t = 0 to t = 5 s.Q5. The displacement x of a particle in meters along the y-axis and timet in seconds alongthex-axis,according to the equation- x= 20m +(12m/s) t(a) Draw a graph if x versus t for t = 0 and t =5 sec(b) What is the displacement come out of the particles initially?(c) What is the slope of the graph obtained?
According to question the data which are given is a = 0.5 m/s2, u =0 at t =0 ,
hence at the end of 5 second the particle has some other velocity and it must have covered some distance or displacement (since the particle is going straight)
using equation of motion
v = u+at
v = 0+0.5 x5 = 2.5 m/s.
now the displacement at the end of 5 second is
s = ut +0.5 at2
s = 0+ 0.5 x 0.5 x 52
= 6.25 metre.
the figure is given below
here the two above figures shows the the velocity and position time graph.
now the equation is given
x = 20+12t
at t =0 x = 20m and at t = 5 x = 20+12x5 = 20+60 =80m
hence the graph is given below
hence initial displacement is 20m
and the slope = tanθ = x/t = 60/5 = 12 .
hence at the end of 5 second the particle has some other velocity and it must have covered some distance or displacement (since the particle is going straight)
using equation of motion
v = u+at
v = 0+0.5 x5 = 2.5 m/s.
now the displacement at the end of 5 second is
s = ut +0.5 at2
s = 0+ 0.5 x 0.5 x 52
= 6.25 metre.
the figure is given below
here the two above figures shows the the velocity and position time graph.
now the equation is given
x = 20+12t
at t =0 x = 20m and at t = 5 x = 20+12x5 = 20+60 =80m
hence the graph is given below
hence initial displacement is 20m
and the slope = tanθ = x/t = 60/5 = 12 .