A bus of mass 5000 kg starts from a rest and rolls down a hill.If it travels a distance of 200 m in 10 seconds.Find
(a) acceleration of a bus
(b) The force acting on the bus
Answers
Answered by
20
Initial velocity (u) = 0 m/sec
Time (t) = 10 sec
Distance (S) = 200 m
Formula used :
S = ut + 1/2 at²
200 = 0×t + 1/2 a×(10)²
200 = 1/2 a×100
100×a = 400
a = 400/100
a = 4 m/sec²
Force = Mass × Acceleration
= 5000 × 4
= 20000 Newton.
Time (t) = 10 sec
Distance (S) = 200 m
Formula used :
S = ut + 1/2 at²
200 = 0×t + 1/2 a×(10)²
200 = 1/2 a×100
100×a = 400
a = 400/100
a = 4 m/sec²
Force = Mass × Acceleration
= 5000 × 4
= 20000 Newton.
Answered by
0
Answer:
ANSWER = 20,000 N
Explanation:
Given that
Mass = 5000kg
Initial velocity =0
Distance= 200m
Time= 10 sec
Then find force acting on it.
We know that
F=ma.
From 1st law of motion
S=ut +1/2at2
200=0+1/2a×100
400=100a
a=4m/s2
Now from equation (i)
F= 5000 × 4
F= 20000 N
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