the velocity of a train is reduced uniformly from 40 km-sec to 30 km-sec while travelling a distance of 100 km
[a] calculate the retardantion
[b] how much farther will it travel befor coming to rest assuming the same retardation
Answers
Answered by
15
Hi dear !!!
here is your answer !
retardation = simply=> ( -ve of acceleration )
v² = u² + 2as
30² = 40² + 2a 100
900-1600 = 200a
- 700 = 200a
a = - 3.5
b ) v² = u² + 2as
0² = 30² + 2× -3.5 × s
- 900 = -7 s
s = 128.57 m
hope it help you dear!!!!
thanks !!!
here is your answer !
retardation = simply=> ( -ve of acceleration )
v² = u² + 2as
30² = 40² + 2a 100
900-1600 = 200a
- 700 = 200a
a = - 3.5
b ) v² = u² + 2as
0² = 30² + 2× -3.5 × s
- 900 = -7 s
s = 128.57 m
hope it help you dear!!!!
thanks !!!
Answered by
18
here initial velocity, u = 40km/sec
and final velocity, v = 30km/sec
and distance travelled = 100km
a. So, using the formula, v² - u² = 2aS
we have, 900 - 1600 = 2a×100
which gives a = -3.5km/sec²
i.e retardation is 3.5km/sec²
b. final velocity, v = 0
initial velocity, u = 30km/sec
Applying the same formula, we get
2aS = v² - u²
2(-3.5)S = 0 - 900
S = 128.57 km
ibrahimsuleman2008:
write down the equations of motion for a body falling freely under gravity a bomb dropped from a balloon reaches the ground in 30sec determine the height of the balloon
[a] if it is at rest in the air [b] if it is ascending with a speed of 100cm when the bombis dropped
For the first part...
S = ut + 1/2 a t^2 i.e h = 0t + (1/2)(9.8)(30)^2 = 4410m = 4.41km
For the second part...
u = 100cm/s = 1m/s, so, h = ut +1/2at^2, h = 1(30) + (1/2)(9.8)(30)^2 = 4440m = 4.444km
may be just 4.44km for the second part :)
Similar questions