A 1000kg vehicle moving with a speed of 20m/s is brought to rest in a distance of 50 meters:
(i) Find the acceleration
(ii) Calculate the unbalanced force acting on the vehicle.
A 1000kg vehicle moving with a speed of 20m/s is brought to rest in a distance of 50 meters:
(i) Find the acceleration
(ii) Calculate the unbalanced force acting on the vehicle.
Vehical Mass = 1000 kg
Initial Velocity (that is speed) let it be u = 20 ms-1
Final Velocity let it be v= 0ms-1
Distance travelled = 50 m
1) let the acceleration be "a"
From equation => v2 = u2 + 2as
putting all value
(0m/s)2 = (20m/s)2 + 2*a*50
0m2/s2 = 400m2/s2 + 100a
0m2/s2 - 400m2/s2 = 100a (Side changing)
-400m2/s2 = 100a meter
-400m2/s2 / 100 m = a (side changing and cancelling m with m2)
-4m/s2 = a
a = -4 m/s2
This shows that the acceleration is in opposite direction that means it is retardation.
retardation = 4m/s2
2)
The unbalance force is
F= mass * acceleration
F= 1000 kg * -4m/s2 ( force is acting on opposite side of initial velocity)
F = -4000 Kgm/s2
F = -4000 N
Vehical Mass = 1000 kg
Initial Velocity (that is speed) let it be u = 20 ms-1
Final Velocity let it be v= 0ms-1
Distance travelled = 50 m
1) let the acceleration be "a"
From equation => v2 = u2 + 2as
putting all value
(0m/s)2 = (20m/s)2 + 2*a*50
0m2/s2 = 400m2/s2 + 100a
0m2/s2 - 400m2/s2 = 100a (Side changing)
-400m2/s2 = 100a meter
-400m2/s2 / 100 m = a (side changing and cancelling m with m2)
-4m/s2 = a
a = -4 m/s2
This shows that the acceleration is in opposite direction that means it is retardation.
retardation = 4m/s2
2)
The unbalance force is
F= mass * acceleration
F= 1000 kg * -4m/s2 ( force is acting on opposite side of initial velocity)
F = -4000 Kgm/s2
F = -4000 N
