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 =
]Create a free account to view solution
View Solution FreeMore Kinematics Questions
A small sphere of radius 'r' falls from rest in a viscous liquid. As a result, heat is produced due to viscous f...A small object of uniform density rolls up a curved surface with an initial velocity v. It reaches up to a maximum heigh...Which the following functions of A and B may be performed if A and B possess different dimensions...A man moves in x - y plane along the path shown. At what point is his average velocity vector in the same direction as h...The velocity of projection of a projectile is (6î + 8ĵ) ms-1. The horizontal range of the projectile is...