Motion in A Straight Line Test - 53

Given,

Horizontal velocity, $${{v}_{x}}=20\,m{{s}^{-1}}$$

Vertical initial velocity, $${{u}_{y}}=25\,m{{s}^{-1}}$$

Acceleration of gravity, $${{a}_{y}}=-10\,m{{s}^{-2}}$$

Apply equation of kinematic

$$ {{v}_{y}}={{u}_{y}}+{{a}_{y}}t $$

$$ {{v}_{y}}=25-10\times 2=5\,m{{s}^{-1}} $$

Hence, velocity $$20\,m{{s}^{-1}}$$ horizontal and $$5\,m{{s}^{-1}}$$ upward

Given that,

Mass $$m=2\,kg$$

Initial velocity $$u=0\,m/s$$

Final velocity $$v=20\,m/s$$

Time $$t=4\,s$$

Now, the acceleration is

  $$ v=u+at $$

 $$ a=\dfrac{v-u}{t} $$

 $$ a=\dfrac{20}{4} $$

 $$ a=5\,m/{{s}^{2}} $$

Now, the force is

  $$ F=ma $$

 $$ F=2\times 5 $$

 $$ F=10\,N $$

Now, the displacement is

  $$ s=ut+\dfrac{1}{2}a{{t}^{2}} $$

 $$ s=0+\dfrac{1}{2}\times 5\times 16 $$

 $$ s=40\,m $$

Now, the work done is

  $$ W=F\centerdot s $$

 $$ W=10\times 40 $$

 $$ W=400\,J $$

Now, the power is

  $$ P=\dfrac{W}{t} $$

 $$ P=\dfrac{400}{2} $$

 $$ P=200\,watt $$

Hence, the power is $$200\ watt$$ 

Given that,

Acceleration $$a=2\,m/{{s}^{2}}$$

Retardation $$a=-3\,m/{{s}^{2}}$$

Time $$t=10\,s$$

Now, from equation of motion

The maximum velocity attained be v at t  

  $$ v=u+at $$

 $$ v=0+2\times t $$

 $$ v=2t\,m/s.....(I) $$

Now, again it reaches velocity 0 after 10 s with retardation

  $$ v=u+at $$

 $$ 0=v-3\left( 10-t \right) $$

Now from equation (I)

  $$ 2t-3\left( 10-t \right)=0 $$

 $$ 2t-30+3t=0 $$

 $$ 5t=30 $$

 $$ t=6\,s $$

Now, put the value of t in equation (I)

  $$ v=2t $$

 $$ v=2\times 6 $$

 $$ v=12\,m/s $$

Hence, the maximum speed is $$12\ m/s$$

Hint: Apply kinematical equations.

Correct Option: Option A

Explanation for correct answer:

Step 1: Find the length of train in terms of velocity and acceleration.

• The displacement of the train is given by:

Given that,

Acceleration, $$a=\frac{dv}{dt}=4-2t$$

At t = 0 s , $$a=4m/{{s}^{2}}$$

At t = 1 s, $$a=2\,m/{{s}^{2}}$$

At t = 2 s, $$a=0$$

At t = 3 s, $$a=4-6=-2\,m/{{s}^{2}}$$

Hence, the acceleration of the body is initially positive but after some time it becomes negative.

Given that,

Height $$h=5\,m$$

$$g=10\,m/{{s}^{2}}$$

Now, for first drop

  $$ {{s}_{1}}=ut+\dfrac{1}{2}g{{t}^{2}} $$

 $$ 5=0+5{{t}^{2}} $$

 $$ t=1\,\sec  $$

It means that the third drop leaves after one second of the first drop. Or, each drop leaves after every $$0.5\ sec$$

Now, distance covered by the second drop in $$0.5\ s$$

  $$ {{s}_{2}}=ut+\dfrac{1}{2}g{{t}^{2}} $$

 $$ s=0+\dfrac{1}{2}\times 10\times 0.5\times 0.5 $$

 $$ s=1.25\,m $$

 Therefore, the distance of the second drop above the ground

  $$ s={{s}_{1}}-{{s}_{2}} $$

 $$ s=5-1.25 $$

 $$ s=3.75\,m $$

Hence, this is the required solution