Kinematics worksheets

Now first step to attempt such question is to visualize the whole process. Here the bus is moving along a straight line and with uniform acceleration
Now what we have
Initial velocity=0
Acceleration=2m/sec²
Now since it is uniform motion we can use given motion formula's in use
$v=u+at$
$s=ut+ \frac {1}{2}at^2$
$v^2=u^2+2as$

a. Distance (s)=?
time(t)=10 sec
So here the most suitable equation is
$s=ut+ \frac {1}{2}at^2$
Substituting given values
$s=(0)10 + \frac {1}{2} 2(10)^2=100 m$

b. Velocity (v) =20 m/s
time (t)=?
So here the most suitable equation is
$v=u+at$
$20=0+2t$
or t=10 sec

c.distance(s)=1.6km=1600m
t=?
So here the most suitable equation is
$s=ut+ \frac {1}{2}at^2$
$1600=\frac {1}{2}(2)t^2$
or t=40 sec

Now first step to attempt such question is to visualize the whole process. Here the object is moving along a straight line and its motion is described by the given equation
Now we
$x=at+bt^2 + ct^3$
Now since its motion is described by the given equation, following formula will be useful in determining the values
$x=\int vdt$
$v=\int adt$
$v= \frac {dx}{dt}$
$a=\frac {dv}{dt}= \frac {d^2x}{dt^2}$
$a=v \frac {dv}{dx}$

a. x=? t=1sec
Here by substituting t=1 in given equation we get the answer
$x=a(1)+b(1)^2+c(1)^3$
$x=a+b+c$ m

b. v=? t=0,v=? t=1
Here we are having the displacement equation, so first we need to find out the velocity equation
So here the most suitable formula is
$v= \frac {dx}{dt}$
or $v=\frac {d}{dt}(at+bt^2+ct^3)$
or $v=a+2bt+3ct^2$
Substituting t=0 we get
$v=a$ m/s
Substituting t=1 we get
$v=a+2b+3c$ m/s

c. a=? t=0,a=? t=1
Now we are having the velocity equation, we need to first find the acceleration equation.
So here the most suitable formula is
$a=\frac {dv}{dt}$
or $a=\frac {d}{dt}(a+2bt+3ct^2)$
$a=2b+6ct$
Substituting t=0 we get
$a=2b$ m/s²
Substituting t=1 we get
$a=2b+6c$ m/s²

Let take the origin at the initial position of the object and upward direction as the positive axis. Let's also take time t as time started when the first object was thrown upward.
Now since it is uniform motion we can use given motion formula's in use
$v=u+at$
$s=ut+ \frac {1}{2)}at^2$
$v^2=u^2+2as$

So from equation of motion
First object
$v_1=v_0 - gt$ and $y_1= v_0t- \frac {1}{2}(gt^2)$ ---(1)
Similarly for the second object. We are writing the equation in same time t variable
$v_2=v_0 - g(t-2)$ and $y_2= v_0(t-2)- \frac {1}{2}g(t-2)^2$ ---(2)
Now when they meet each other, then
$y_1= y_2$
$v_0t- \frac {1}{2}(gt^2) = v_0(t-2)- \frac {1}{2}g(t-2)^2$
Solving it
$t= \frac {v_0}{g}+1=5.09$sec
Substituting this value in any above equation will give the value of height
$H=40 \times 5.09- \frac {1}{2} (9.8 \times 5.09 \times 5.09) =76.6$ m
The respective velocities can be found by using the first set of each equation 1 and 2
For First object
$v_1=v_0 - gt =40-(9.8 \times 5.09)=-9.882$ m/sec
For second object
$v_2=v_0 - g(t-2) =40-9.8(5.09-2)=9.718$ m/sec

a. Equation of motion is given by
$x=pt(1-qt)$
Velocity = $\frac {dx}{dt}$
So $v=p(1-2qt)$
Acceleration = $\frac {dv}{dt}$
So $a=-2pqt$
Hence,
Velocity vector= $p(1-2qt) \mathbf{i}$
Acceleration vector=$-2pqt \mathbf{i}$

b. Equation of motion is given by
$x=pt(1-qt)$
at t=0,x=0 and $t= \frac {1}{q}$ ,x=0
So at t=1/q ,it will reach it initial position
Equation of velocity
$v=p(1-2qt)$
From the equation we can see that
For $ t < \frac {1}{2q}$ , $v>0$ so speed=$|v|= p(1-2qt)$
At $t=\frac {1}{2q}$ ,$v=0$
For $t > \frac {1}{2q}$ ,$v < 0$ so speed=$|v|=p(2qt-1)$
Distance traveled from time t=0 to t=1/2q
$s_1=\int_{0}^{\frac {1}{2q}} p(1-2qt)dt = \frac {p}{4q}$
Distance traveled from time t=1/2q to t=1/q
$s_2= \int_{\frac {1}{2q}}^{\frac {1}{q}} p(2qt -1)dt = \frac {p}{4q}$
So total distance traveled=$ \frac {p}{4q} + \frac {p}{4q} =\frac {p}{2q}$

Since balloon start's from rest, the velocity of the balloon when it risen to height h is given by
$v^2= u^2+2gH$
$v^2=0 + 2 \times \frac {g}{8} \times h$
$v=\frac {1}{2} \sqrt {gh}$
When the object is dropped, it has the velocity this same velocity in upward direction. Taking upward direction as positive,
$-h= \sqrt {gh} t- \frac {1}{2} gt^2$
Where h is negative as we have taken upward as positive axis , t is the time taken taken to reach the ground and acceleration is -g as gravity is active downwards
Equation can be written in the form
$gt^2-\sqrt {gh} t-2h=0$
Solving this equation and taking positive root
$t=2 \sqrt {\frac {h}{g}}$

Speed of the Police Motorcycle w.r.t ground = $v_m$
Speed of the Bullet w.r.t to Motorcycle or Muzzle= $v_b$
The speed of the bullet w.r.t ground=Speed of the bullet w r.t to Motorcycle + speed of the motorcycle w.r.t to ground
So Speed of the bullet w.r.t to ground= $v_b + v_m$

Speed with bullet hit the thief=Speed of bullet w.r.t to ground - Speed of Thief w.r t ground
= $v_b + v_m- v_t$

Speed of the bullet w.r t another motorcycle=Speed of the bullet w.r.t ground - Speed of the Motorcycle
= $v_b + v_m - v$

a. Here the nut initially has the velocity of the elevator when it fell off. Let take upward direction as the positive direction
Then $v_0=3 $m/s at t=0 and $a=-g=-9.8$ m/s²
Now time to hit the ground t=2
So
$H= v_0t+ \frac {1}{2}gt^2$
$H=3 \times 2 + \frac {1}{2} \times (-9.8) \times 2 \times 2 =-13.6$ m
So the bottom of the shaft was 13.6 m below when the nut fell off from elevator.
So elevator was 13.6 m above the shaft when the nut fell off

b. Now Let calculate the nut displacement at t=.25
$H= v_0t+\frac {1}{2}gt^2$
Substituting the values as same above
H=.44m
Thus the nut was above its starting point. This makes sense if we remember the initial velocity as upward.
So the total height above the bottom will be
=.44+13.6=14 m

c. Now the elevator perspective
v=3m/sec upward
So it will move 6m upward in 2 sec.
So total height above the ground will be
=6+13.6
=19.6 m

d. $v^2=u^2+2ah$
From nut perspective
u=3m/sec
a=-g=-9.8m/s²
at v=0
$u^2=2gH$
or H=3*3/2*9.8
=.45m
So Total height above the bottom of shaft=13.6+.45=14.1m

e. Total distance traveled by nut=.45+.45+13.6=14.5m

Given $x=- \frac {2}{3}t^2 +16t+2$ (1)
Velocity=$ \frac {dx}{dt}=\frac {-4}{3} t+16$
so velocity at t=0 =16
and velocity at t=1 =$\frac {-4}{3} +16=\frac {44}{3}$
Acceleration is given as =$ \frac {d^2x}{dt^2}=- \frac {4}{3}$
so acceleration is time independent and it is constant
Displacement at t=0 can be found by simply substituting the values of t=0 in equation (1)
So, Displacement at(t=0)=2
Now $v=- \frac {4}{3}t+16$
when v=0
then $t= \frac {4}{3}$ sec
Displacement can be found by substituting the value of t=4/3 in equation (1)
$=- \frac {32}{27} + \frac {64}{3} +2$
$= \frac {(-32+576+54)}{27}=\frac {598}{27}$ m

At t=0 let man's position be the origin $x_{m0}=0$.The bus door is then at $x_{b0}=6$ m
The equation of motion for the man and the bus
$x_m=x_{m0}+v_{m0}+ \frac {1}{2}a_m t^2$
$x_b=x_{b0}+v_{b0}+\frac {1}{2}a_b t^2$
Now given are
$v_{m0}=4$ m/s , $v_{b0}=0$ ,$a_m=0$ , $a_b=1.2$ m/s²
Thus
$x_m=4t$
$x_b=6+.6t^2$
When the man catches the bus
$x_m=x_b$
so $ 4t=6+.6t^2$
or $3t^2-20t+30=0$
Solving the Quadratic Formula
t= 2.3 and 4.4 s
Note here are two positive solutions .This can be explained as. The first time $t_1=2.3$ s, corresponds to his first reaching the door. This is the real answer. However equation we have solved doesn't know he will stop running and board the bus, the equation have continuing up a running at constant speed. He thus goes on past the bus; but since the bus is accelerating, it eventually builds a larger velocity then the man and will catch with him at 4.4 sec
if the initial position of the bus is 10 m then $x_m=x_b$
gives $3t^2-20t+50=0$ which has only complex roots so no real time exits at which man catches up the bus