Question

At the height of 75 m, a particle A is thrown up with V = 10 m/s and B particle is thrown with V = 10 m/s and C particle is released with V = 0 m/s. Draw graphs of each particle.
a. Displacement - time
b.Speed - time
c. Velocity - time
d. Acceleration - time

Answer 1

Taking origin at the top of the building and upward direction positive.

For particle A

u = + 10m/s a=-10 m/s
Using y = ut +(1/2) a$t^2$
= 10 t - (1/2) x 10 x $t^2$ = 10t - 5$t^2$ ...(i)
Equation (i) is the equation of parabola open down.
The parabola passes origin when y = 0, hence from (i)
0 = 10t - 5$t^2$
t = 0,2 s
For velocity- time relation, v = u + at
v = 10 - 10t ..(ii)
This relation is straight line relation with negative slope and positive intercept.
At maximum height, velocity of the particle is zero.
0= 10-10 t => t = 1 s
Maximum height from (i), $y_{max}$ = 10 x 1 - 5 x 1 x 1 = 5 m/s
It will reach ground where y=-100 m
Using (i) again, -75 = 10t -5$t^2$ => (t - 5) (t + 3) = 0
When given t = 5 s, i.e., the particle will hit ground at t = 5 s.
Velocity at the time of hitting
Using (ii), v = 10 -10 x 5 = 40 m/s
For speed-time relation, the speed is always positive the mirror-image of velocity-time relation on the positive side.
For drawing an acceleration-time graph, we have taken downward direction as negative and acceleration is constant throughout the motion of the particle, i.e., -10 $m{s}^{-2}$. Hence, the acceleration time graph will be straight line parallel with time axis and below the time axis.

For particle B:

u = -10 m/s a = -10$m{s}^{-2}$
Displacement-time relation,
y = 10t - (1/2) x 10 x $t^2$ = -10t - 5$t^2$ ...(iii)
It is a parabola open down and it passes through origin.
The particle will reach ground when y = -75 m From (iii) -75 = -10t - 5$t^2$
or $t^2$ + 2f+15 = 0 => t = 3s
The velocity of the particle in relation with time
v = 0 - 10t => v = -10t ...(iv)
This is a straight line with negative slope and passing through origin. The particle reaches ground at t = 3 s, velocity of the particle when-it hits ground is v = -10 x 3 = -30 m/s

For particle C:

u = 0, a = -10 $m{s}^{-2}$
Using y = ut + (1/2)a$t^2$
= 0 -(1/2)(-10)$t^2$ = -5$t^2$ ...(v)
It is a parabola open down and it passes through origin.
For velocity-time relation v = u + at
v = 0- 10t =-10t
Straight line with negative slope passes through origin.
It will reach round at y = -75 m
-75 = -5$t^2$
t = $\sqrt{15}$s
Velocity at the time when it reaches ground v = -10$\sqrt{15}$s.