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Question: A truck starts from rest and accelerates uniformly at \(2 \mathrm {~ms} ^ { - 2 }\) . At t = 10 s, ...

A truck starts from rest and accelerates uniformly at 2 ms−22 \mathrm {~ms} ^ { - 2 } . At t = 10 s, a stone is dropped by a person standing velocity of the stone at t = 11 s is (Take g=10 ms−2\mathrm { g } = 10 \mathrm {~ms} ^ { - 2 } )

A

5 ms−1\sqrt { 5 } \mathrm {~ms} ^ { - 1 }

B

105 ms−110 \sqrt { 5 } \mathrm {~ms} ^ { - 1 }

C

20 ms−1\sqrt { 20 } \mathrm {~ms} ^ { - 1 }

D

55 ms−15 \sqrt { 5 } \mathrm {~ms} ^ { - 1 }

Answer

105 ms−110 \sqrt { 5 } \mathrm {~ms} ^ { - 1 }

Explanation

Solution

For truck u=0,a=2ms−2,t=10su = 0 , a = 2 m s ^ { - 2 } , t = 10 s

Let v be the velocity of the truck after 10s

As v=u+atv = u + a t

∴v=0+(2ms−2)(10s)=20ms−1\therefore v = 0 + \left( 2 m s ^ { - 2 } \right) ( 10 s ) = 20 m s ^ { - 1 }

For stone,

When the stone is dropped from the moving truck it will possess a velocity along the horizontal equal to that of the truck at that time

∴vx=v=20 ms−1\therefore v _ { x } = v = 20 \mathrm {~ms} ^ { - 1 }

Initial velocity of the stone along the vertical is zero. Let vyv _ { y } be the velocity of the stone along the vertical at t=11st = 11 s i.e. after being dropped out of the truck

∴vy=0+(10 ms−2)(1s)=10 ms−1\therefore v _ { y } = 0 + \left( 10 \mathrm {~ms} ^ { - 2 } \right) ( 1 s ) = 10 \mathrm {~ms} ^ { - 1 }

The resultant velocity of the stone at t=11 s is

vR=vx2+vy2=(20ms−1)2+(10ms−1)2v _ { R } = \sqrt { v _ { x } ^ { 2 } + v _ { y } ^ { 2 } } = \sqrt { \left( 20 m s ^ { - 1 } \right) ^ { 2 } + \left( 10 m s ^ { - 1 } \right) ^ { 2 } }

=500 ms−1=105 ms−1= \sqrt { 500 } \mathrm {~ms} ^ { - 1 } = 10 \sqrt { 5 } \mathrm {~ms} ^ { - 1 }