Physics Solution of Distance Travelled

Physics

1.1:- Solution:-
Distance travelled = s in meter
Time Taken = t in sec.
u is the initial velocity = 5 m/s
a is the acceleration = 3 m/s2
t= 10 seconds
Second equation of the motion
S= ut + ½ a t2……………….(1)
On putting the value of u, a and t in equation (1) we can calculate the distance in meter.
S= 5 x 10 +1/2(3 x(10 x 10))
S=5 x 10 + ½ (300)
S= 50 + 150
S= 200 meters
Graph between the distance (S) and time (t)

The Graph is linear. It means that distance and time changes uniformly. It travels same interval in same
distance.
b)-1 Solution:-
S= ut + ½ at2
On differentiating
ds/dt =u+1/2(at2)
ds/dt = u+at
Given: v=ds/dt
Therefore
V= u + at…………………………..First equation of the Motion
b)-2 Solutions:-

a =dv/dt …….given
a =dv/dt= d2s/dt2
a= dv/dt = d/dt (u+at)
On differentiating
It becomes
dv/dt= d/dt (0+a)
a=acceleration
1.3 Solutions:-
u is the initial velocity = 5 m/s
a is the acceleration = 3 m/s2
for calculate the value velocity
at t= 10 seconds
by using ist equation of the motion
v= u + at…………………………….(2)
On putting the value in equation (2)
We obtained
V= 5 + 3 x 10
V= 5+30
V= 35 m/s
Acceleration (a) change in velocity/time
From the above graph
a= (Final velocity – initial velocity)/time
Final velocity= 35m/s
Initial velocity = 3 m/s
a= (35-3)/10

a=3.2 m/s2
2 Solutions:-
Displacement of mass = y= 2x2+4x3
Steps to find the turning Points:-
Step 1:- find dy/dx
Step 2:- set dy/dx=0 solve for x values
Step 3:-Substitute the value of x in the original y= f(x) and get value of y
Step 4:-Write in form (x,y)
y= 2x2+4x3
On differentiating
It becomes
dy/dx= 4x+12x2
4x=-12x2
x=-1/3
Substitute the value of x in the original y= f(x)
y= 2(1/3)2+4(1/3)3
y= 2/9-4/27
y=2/27
Turning points: (x, y) = (-1/3, 2/27)
3 Solutions:-
Equation for the Charging of Capacitor
V= v0 e-t/T …………………………………(4)
T= is Time constant of the circuit equal to RC

R – Resistance of the circuit
C- Capacitor of the Circuit
v0- Initial voltage
V0 = 11
For t1 = 3 sec
On putting the value of t1 in equation (4)we obtained
V= v0 e-3/T
By using chain rule
dv/dt = v0 d/dt e-3/T + e-3/T d/dt v0
Differentiation of constant term is zero
dv/dt = v0 d/dt e-3/T
dv/dt = 11 (-1) e1/T
For t1 = 5 sec
V= v0 e-5/T
By using chain rule
dv/dt = v0 d/dt e-5/T + e-5/T d/dt v0
Differentiation of constant term is zero
dv/dt = v0 d/dt e-5/T
dv/dt = 11(-1) e1/T
(b) Second derivative
dv/dt = v0 (-1) e1/T
On differentiating again
d2v/dt2=0

4 Solutions:-
(a) V = 12(1- e-t/2 )
On differentiating
dv/dt = 12- 12e-t/2
dv/dt = 0 – 12(-1) e-t/2
dv/dt = 0 – 12(-1) e1/2
e1/2 = 1.6487
dv/dt = 12 x 1.6487
dv/dt = 19.784
(b) When t= 5 sec
dv/dt = 12- 12e-t/2
dv/dt = 12e-5/2
dv/dt = 12 x 12.18
dv/dt = 146.18
When t= 3sec
dv/dt = 12- 12e-t/2
dv/dt = 12e3/2
dv/dt = 12 x 4.481
dv/dt = 53.78
(c) Second derivative of dv/dt
dv/dt = 0 – 12(-1) e-1/2
d2v/dt2= 12
5 Solutions:-

d/dx logax = 1/x [Ln(a)]
Assumption: log is equal to log10
By using the Chain Rule
d/dt (u,v) =- u*dv/dt+ v*du/dt
G = 20 log10V
a) dG/dV = [20/ [10V ln 10)] x10
= 20 / (V ln 10)
dG/dV = 20 / (V ln 10)
b) d2G/dV2= ( 20 / ln 10 )(-1/ v2)
= -20 (20 /ln 10 )/v2
d2G/dV2= -20 (20 /ln 10 )/v2
6 Solutions:-
Y= 2e-t sin3t
Product rule for differentiation: d/dt (u,v) =- u*dv/dt+ v*du/dt
V=dy/dt = 2[e-t sin3t]
V=dy/dt =2{[e-t d/dt(sin3t)]+ sin3t d/dt (e-t)}
V=dy/dt = 2 [e-t x3x (cos3t) + sin 3t (-1) e-t )
V=dy/dt = 2[3 e-t cos3t – e-t sin 3t]
V=dy/dt = 6 e-t cos 3t – 2 e-t sin3t
7 Solutions:-
Velocity v= (2t+3)4
By chain rule d/dt (u,v) =- u*dv/dt+ v*du/dt
v= (2t+3)4 *1

dv/dt = 4(2t+ 3)3 +d/dt (2t+3)
dv/dt = 8 (2t +3)3
a=d/dt (dv/dt)
Therefore
a=d/dt (dv/dt) = 8 x 3(2t+3)2 d/dt (2t+3)
= 24 (2t+3)2 x 2
= 48 (2t + 3)2
= 48 (4t2 +12t +9)
a=d/dt (dv/dt) = 48 (4t2 +12t +9)
8 Solutions:-
Y = sint/t
By chain rule d/dt (u,v) =- u*dv/dt+ v*du/dt
Y = (t)-1 sint
U= (t)-1 and v= sint
On differentiating
It becomes
dy/dt = (t)-1 cost + sint (-1)t-2
dy/dt = cost/t- sint /t2
dy/dt =(tcost –sint)/t2
dy/dt =(tcost –sint)/t2
9 Solutions:-
Perimeter = 2 (l + b)
Perimeter = 2 (l + b) = 750
(l + b) = 375

l= x
Therefore
b= (375 – x)
Area will be Maximum when dA /dx= 375-2X=0
2x = 375
x= 375/2
A = 375 x 375/4
A= 35, 156.251 m2
10 Solutions:-
Squares cut from the corners are x×x inches
Open-top box height = x mm
Width of 12−2h mm
length = 120 mm
New length = 120-2x mm
Width = 140 mm
New Width = 140 -2x mm
so it's volume will be
V(x) = x (120−2x) (140−2x) mm3
On solving above equation it becomes
V(x) =x (120x -2x)(140-2x)
V(x) =x (120 (140-2x) -2x (140 -2x))
V(x) =x (16800 -24 x – 280 x +4x2)
V(x) =x (4 x2 – 304 x -16800)
V(x) = 4x3 -304x2 -16800

Critical points occur when the derivative (dv/dx) is equal to zero.
Therefore dv/dx =0
dv/dx = 12x2-608x
dv/dx = x (12x- 608)
dv/dx = 4x (3x-152)
x= 50.6666
x=51 mm
(b) So its volume will be
x= 51mm would result in widths and lengths of zero, and therefore a volume of zero), so it is obviously
not the critical point for the maximum.
(c) Therefore the maximum volume is achieved by cutting out squares that are
51 x 51 inches from the corners of the larger sheet.