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GATE Civil Engineering (CE) 2018 Shift 2 Solved Paper
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© examsnet.com
Question : 41 of 65
Marks:
+1
,
-0
A car follows a slow moving truck (travelling at a speed of
10
 
m/s
10\,\text{m/s}
10
m/s
) on a two-lane two-way highway. The car reduces its speed to
10
 
mm/s
10\,\text{mm/s}
10
mm/s
and follows the truck maintaining a distance of 16
 m
\ \text{m}
Â
m
from the truck. On finding a clear gap in the opposing traffic stream, the car accelerates at an average rate of
4
 
m/s
2
4\,\text{m/s}^{2}
4
m/s
2
, overtakes the truck and returns to its original lane. When it returns to its original lane, the distance between the car and the truck is
16
 
m
.
16\,\text{m}.
16
m
.
The total distance covered by the car during this period (from the time it leaves its lane and subsequently returns to its lane after overtaking) is
64
 
m
64\,\text{m}
64
m
72
 
m
72\,\text{m}
72
m
128
 
m
128\,\text{m}
128
m
144
 
m
144\,\text{m}
144
m
Validate
Solution:
Overtaking time,
T
=
  
4
s
a
=
  
4
×
16
4
=
4
 
s
T=\sqrt{\;\frac{4s}{a}}=\sqrt{\;\frac{4 \times 16}{4}}=4\,\text{s}
T
=
a
4
s
​
​
=
4
4
×
16
​
​
=
4
s
S
=
S=
S
=
Space heady way
=
16
 
m
=16\,\text{m}
=
16
m
a
=
a=
a
=
Acceleration
=
4
 
m/s
2
=4\,\text{m/s}^{2}
=
4
m/s
2
Distance travelled by vehicle
=
S
2
=S_{2}
=
S
2
​
S
2
=
u
T
+
  
1
2
a
T
2
=
10
×
4
+
  
1
2
×
4
×
4
2
=
72
 
m
S_{2}=u T+\;\frac{1}{2}a T^{2}=10 \times 4+\;\frac{1}{2} \times 4 \times 4^{2}=72\,\text{m}
S
2
​
=
u
T
+
2
1
​
a
T
2
=
10
×
4
+
2
1
​
×
4
×
4
2
=
72
m
© examsnet.com
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