A motorcyclist riding motorcycle A who is travelling at 36 km/h applies the brakes and
stops the motorcycle in 10 s. Another motorcyclist of motorcycle B who is travelling at 18 km/h applies the breaks and stops the motorcycle in 20s. Plot speed-time graph for the two motorcycles.
Which of the two motorcycles traveled farther before it comes to a stop?
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Answered by
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Heyy Mate here is Ur Answer.....
Given :
Case I:
For motorist A,
Initial Velocity= u=36km/h=36x5/18=10 m/s
Time=t=10s
by using third equation of motion:
V²-u²=2as
0-10x10=2s(V-u)/t
0-10x10=2s(0-10)/10
-2s=-100s
=100/2=50m
Case II:
For motorist B,
Initial velocity=18km/h=18x5/18=5m/s
time=t=20s
By using third equation of motion:
V²-u²=2as
V²-u²=2s(v-u)/t
0-5x5=2s(0-5)/20-25=-s/2s=50m
so both A and B cover the same distance.
To draw graph ;we need to find acceleration:
For Motorist A:a=(v-u)/t=(0-10)/10=-1 m/s²
For Motorist B:a=(v-u)/t=(0-5)/20=-0.25 m/s²
Time is from 0 to 20 sec and speed is from 5 to 0.25 m/s
Please refer the attachment for the graphs:
Graph 1- Motorist A
Graph 2- Motorist B
HOpe It Helps U :-)
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Answered by
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Answer:
here is your answer
Explanation:
Given :
Case I:
For motorist A,
Initial Velocity= u=36km/h=36x5/18=10 m/s
Time=t=10s
by using third equation of motion:
V²-u²=2as
0-10x10=2s(V-u)/t
0-10x10=2s(0-10)/10
-2s=-100s
=100/2=50m
Case II:
For motorist B,
Initial velocity=18km/h=18x5/18=5m/s
time=t=20s
By using third equation of motion:
V²-u²=2as
V²-u²=2s(v-u)/t
0-5x5=2s(0-5)/20-25=-s/2s=50m
so both A and B cover the same distance.
To draw graph ;we need to find acceleration:
For Motorist A:a=(v-u)/t=(0-10)/10=-1 m/s²
For Motorist B:a=(v-u)/t=(0-5)/20=-0.25 m/s²
Time is from 0 to 20 sec and speed is from 5 to 0.25 m/s
Please refer the attachment for the graphs:
Graph 1- Motorist A
Graph 2- Motorist B
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