Tag: motion and rest

Questions Related to motion and rest

A boy running at an average speed of $4 km hr^{-1}$ reaches school from his home in $\frac{1}{2}hr$. The distance of the school from his home is

  1. $2 km$

  2. $8 km$

  3. $4 km$

  4. $6 km$


Correct Option: A
Explanation:

Given:$v =4\frac{Km}{h}$
           

     $t =\dfrac{1}{2}$hour
   as $v=\dfrac{Distance}{t}$
        $Distance=4\times\dfrac{1}{2}$
        $Distance=2Km$

If the displacement of an object is zero, the actual distance traversed by it will be :

  1. Zero

  2. Infinite

  3. May not be zero

  4. None of these


Correct Option: C
Explanation:

Displacement is the distance between the initial and final position of an object. When an object moves in a circular path of radius R, its displacement will be zero but the distance traveled by the object will be the perimeter i.e $2\pi R$. 

Thus, if displacement of an object is zero, the distance traversed by it may not be zero.

Sunil and Rohit start at one end of a street, the origin and run to the other end, then head back. On the way back, Sunil is ahead of Rohit. Which statement is correct about the distances run and the displacements from the origin before reaching the origin?

  1. Sunil has run a greater distance and his displacement is greater than Rohit's

  2. Rohit has run a greater distance and his displacement is greater than Sunil's

  3. Sunil has run a greater distance but his displacement is lesser that Rohit's

  4. Rohit has run a greater distance but his displacement is lesser than Sunil's


Correct Option: C
Explanation:

Before reaching origin, distance travelled by Sunil is more than Rahul since he is ahead of Rohit. But displacement of Rohit is more as he is far from the origin than Sunil.

A free falling body travels ___ of total distance in 5th second.

  1. 8 %

  2. 12 %

  3. 25 %

  4. 36 %


Correct Option: D
Explanation:
The correct option is D

Given,

A free falling body 

Distance traveled in 5s

$S=ut+\dfrac{1}{2}gt^2$ since $u=0,g=10m/s$

$=\dfrac{1}{2}\times10\times5^2$

$=12m$

Distance travelled in 5^th sis:

Distance travelled in 5s- Distance travelled in 4s

Distance travelled in $4s=\dfrac{1}{2}\times10\times4^2$

$=80m$

So,

$s _5-s _4=125-80=45$

Therefore,

$\dfrac{s _5-s _4}{s _5}\times100=\dfrac{45}{125}\times100=36\%$

A particle experiences a constant acceleration for $6s$ after starting from rest. If it travels  distance $x _{1}$ in the first two seconds and a distance $x _{2}$ in the next two next seconds and $x _{3}$in the next two seconds the n  $x _{1}: x _{2}; x _{3}::1:3:5.$

  1. True

  2. False


Correct Option: A