Question

A honeybee flies with a velocity of 2 m/s. It turns back and moves with the same velocity. The total time taken to change the direction of velocity is 2 s. The magnitude of acceleration of the bee is *
1 m/s²
3 m/s²
4 m/s²
2 m/s²​

Answer 1

Answer:

If the honeybee has to fly in straight line then he must fly at some angle theta with the AB. And component of velocity of plane in direction of air flow with respect to ground should be zero. i.e. u=vSinθ. From here Sinθ=  

v

u

​  

 

And component of velocity of plane which will help plane to cover the distance l  is vCosθ

So time taken to cover distance l is given by  

vCosθ

l

​  

 and Cosθ=  

1−(  

v

u

​  

)  

2

 

​  

.

So time taken from A to B is  

v  

2

−u  

2

 

​  

 

l

​  

 same will be case while returning so total time will be  

v  

2

−u  

2

 

​  

 

2l

​  

 

So best option is A.

Explanation:

Answer 2

Explanation:

Velocity, u=∣t−2∣

Time, t=4 s

at t=0, u=2 m/s

t=2, u=0 m/s

t=4 ,u=2 m/s

between time t=0 to t=2sec

a=

2

0−2

=−1 m/s

2

s

1

=ut+

2

1

at

2

s

1

=2×2+

2

1

×(−1)×4=2

Between time t=2 to t=4sec

a=

2

0−2

=1 m/s

2

s

2

=

2

1

×1×4=2

Thus,

Total distance;

s

1

+s

2

=2+2=4 m

Thus, during the 4 seconds, the bird travels a distance of 4 m.