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