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  • A car starts with constant acceleration a = 2m/s2  at t = 0. Two coins are released from the car at t = 3 & t = 4. Each coin takes 1 second to fall on ground. Then the distance

A car starts with constant acceleration a = 2m/s2  at t = 0. Two coins are released from the car at t = 3 & t = 4. Each coin takes 1 second to fall on ground. Then the distance (in meter(m)) between the two coins will be : (Assume coin sticks to the ground) 

Option: 1

9


Option: 2

7


Option: 3

15


Option: 4

2


Answers (1)

best_answer

 

1st equation or velocity time equation -

 

V=u+at

V = Final velocity

u = Initial velocity

A = acceleration

T = time

-

 

 

2nd equation or Position- time equation -

s= ut +\frac{1}{2}at^{2}

s\rightarrow Displacement

u\rightarrowInitial velocity

a\rightarrowacceleration

t\rightarrow time

 

-

 

 

v = at = 2t

  Velocity of car

                at t = 3  V1 = 6 m/s              

                at    t = 4    v2 = 8 m/s

 Coin 1 will fall with horizontal velocity 6 m/s & second coin will fall with horizontal velocity 8 m/s. Both will travel 6 m & 8 m horizontally before they fall from the point of release. 

   Car moves \frac{6+8}{2}\times 1 = 7 m.

In fourth second, position of first coin and second coin respectively x_{1} = 6, x_{2} = 7 + 8 = 15 

\Rightarrow x_{2}- x_{1}=15-6=9m

 

Posted by

Riya

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