Answer:
v(t) = 21.3t
v(t) = 5.3t
Explanation:
When no sliding friction and no air resistance occurs:
where;
Taking m = 3 ; the differential equation is:
By Integration;
since v(0) = 0 ; Then C = 0
v(t) = 21.3t
ii)
When there is sliding friction but no air resistance ;
Taking m =3 ; the differential equation is;
By integration; we have ;
v(t) = 5.3t
iii)
To find the differential equation for the velocity (t) of the box at time (t) with sliding friction and air resistance :
The differential equation is :
=
=
By integration
Since; V(0) = 0 ; Then C = -48
Answer:43.34 m
Explanation:
Given
acceleration(a)
Initial Velocity(u)=0 m/s
After 6 s fuel runs out
Velocity after 6 s
v=u+at
After this object will start moving under gravity
height reached in first 6 s
s=36 m
After fuel run out distance traveled in upward direction is
here v=0
u=12 m/s
Answer:
0.8 seconds
Explanation:
F=ma
Let x be the seconds the force is applied.
m = 20kg
F = 50 Newtons (kg*m/sec^2)
acceleration, a, is provided for x seconds to increase the speed from 1 m/s to 3 m/s, an increase of 2m/s
Let's calculate the acceleration of the cart:
F=ma
(50 kg*m/s^2) = (20kg)*a
a = 2.5 m/s^2
---
The acceleration is 2.5 m/s^2. The cart increases speed by 2.5 m/s every second.
We want the number of seconds it takes to add 2.0 m/sec to the speed:
(2.5 m/s^2)*x = 2.0 m/s
x = (2.0/2.5) sec
x = 0.8 seconds
Answer:
F= 9492.12 N
Explanation:
Given that
mass, m = 1598 kg
Initial velocity ,u= 0 km/h
Final velocity v= 93.9 km/h
v =26.08 m/s
Distance ,d= 57.2 m
We know that
v² = u² + 2 a d
a=acceleration
u=initial velocity
v= final velocity
d=distance
26.08²= 0² + 2 x a x 57.2
a= 5.94 m/s²
We know that
Force = mass x acceleration
F= m a
F= 5.94 x 1598 N
F= 9492.12 N