A train moving at 30m/s brakes and decelerate uniformly at 9m/s^2. How much distance does it travel before coming to stop? About the author Eden
Answer Correct option is B 240m Initial velocity u=30m/s Final velocity, v=10m/s Distance traveled, s=240m v 2 −u 2 =2as where ‘a’ is acceleration 100−900=2as a= 2×240 −800 =− 6 10 m/s 2 F=ma=−m 6 10 N If force is increased by 12.5%,new force F ′ =(100+12.5)% ×F =112.5 % of F F F ′ = 100 112.5 =1.125 ma ma ′ =1.125 a ′ =1.125a=1.125× 6 −10 =−1.875m/s 2 New distance traveled when v ′ =0 v ′2 −u 2 =2a ′ s ′Answer Correct option is B 240m Initial velocity u=30m/s Final velocity, v=10m/s Distance traveled, s=240m v 2 −u 2 =2as where ‘a’ is acceleration 100−900=2as a= 2×240 −800 =− 6 10 m/s 2 F=ma=−m 6 10 N If force is increased by 12.5%,new force F ′ =(100+12.5)% ×F =112.5 % of F F F ′ = 100 112.5 =1.125 ma ma ′ =1.125 a ′ =1.125a=1.125× 6 −10 =−1.875m/s 2 New distance traveled when v ′ =0 v ′2 −u 2 =2a ′ s ′ 0−900=2×(−1.875)×s ′ s ′ = 2×1.875 900 =240m 0−900=2×(−1.875)×s ′ s ′ = 2×1.875 900 =240m Reply
Answer
Correct option is
B
240m
Initial velocity u=30m/s
Final velocity, v=10m/s
Distance traveled, s=240m
v
2
−u
2
=2as where ‘a’ is acceleration
100−900=2as
a=
2×240
−800
=−
6
10
m/s
2
F=ma=−m
6
10
N
If force is increased by 12.5%,new force
F
′
=(100+12.5)% ×F
=112.5 % of F
F
F
′
=
100
112.5
=1.125
ma
ma
′
=1.125
a
′
=1.125a=1.125×
6
−10
=−1.875m/s
2
New distance traveled when v
′
=0
v
′2
−u
2
=2a
′
s
′Answer
Correct option is
B
240m
Initial velocity u=30m/s
Final velocity, v=10m/s
Distance traveled, s=240m
v
2
−u
2
=2as where ‘a’ is acceleration
100−900=2as
a=
2×240
−800
=−
6
10
m/s
2
F=ma=−m
6
10
N
If force is increased by 12.5%,new force
F
′
=(100+12.5)% ×F
=112.5 % of F
F
F
′
=
100
112.5
=1.125
ma
ma
′
=1.125
a
′
=1.125a=1.125×
6
−10
=−1.875m/s
2
New distance traveled when v
′
=0
v
′2
−u
2
=2a
′
s
′
0−900=2×(−1.875)×s
′
s
′
=
2×1.875
900
=240m
0−900=2×(−1.875)×s
′
s
′
=
2×1.875
900
=240m