A bus started from bustand at 8.00am, and after staying for 30 minutes at a destination, it returned back to the busstand. The destination is 27 miles from the busstand. The speed of the bus is 18mph. During the return journey bus travels with 50% faster speed.At what time does it return to the busstand?
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Given: s = 27 miles u= 18 mph
v= u + 50%u
= u(1 + 50/100)
= 18(1 + 0.5)
= 18 + 9
= 27 mph
Solution:. while going to destination
t = s/u
= 27/18
= 1.5
While returning back
t`= s/v
= 27/27
=1 hour
Hence, total time taken in travelling
t"= t+t`
= 1.5+1
=2.5 hours
time elapsed on destination
t,=0.5 hour
Total time taken
T= t,+t"
= t,+t"
= 0.5+2.5
=3 hours
Starting time- 8:30 a.m.
Time gap = 3 hours
hence,
Ending time- 11:30 a.m.
hence the bus returned at 11:30 on the bus stand
Easy, isn't it?
v= u + 50%u
= u(1 + 50/100)
= 18(1 + 0.5)
= 18 + 9
= 27 mph
Solution:. while going to destination
t = s/u
= 27/18
= 1.5
While returning back
t`= s/v
= 27/27
=1 hour
Hence, total time taken in travelling
t"= t+t`
= 1.5+1
=2.5 hours
time elapsed on destination
t,=0.5 hour
Total time taken
T= t,+t"
= t,+t"
= 0.5+2.5
=3 hours
Starting time- 8:30 a.m.
Time gap = 3 hours
hence,
Ending time- 11:30 a.m.
hence the bus returned at 11:30 on the bus stand
Easy, isn't it?
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