KinematicsHard
Question
A 2m wide truck is moving with a uniform speed of 8 m/s along straight horizontal road. A pedestrian starts crossing the road at an instant at road at an instant when the truck is 4 m away from him. The minimum constant constant velocity when which he should run to avoid an accident is : -
Options
A.1.6 √5 m/s
B.1.2 √5 m/s
C.1.2 √7 m/s
D.1.6 √7 m/s
Solution
Time to cross 2m is 
.................
To avoid an accident
Displacement = 4 + v cos θ ×
8 ×
= 4 + cot θ
v sin θ =
v min =
= 1.6 √5m/s
[∵ (a cos θ + b sin θ) has max. value =
]
.................To avoid an accident
Displacement = 4 + v cos θ ×
8 ×
= 4 + cot θv sin θ =
v min =
= 1.6 √5m/s[∵ (a cos θ + b sin θ) has max. value =
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