Physics, asked by prishakunwar, 1 month ago

A train starts from rest and accelerates uniformly at the rate of 5 m/s2for 5 sec. Calculate the velocity of the train and the distance covered by the train after 5 sec.

Answers

Answered by taslimuddinahmad60
28

Refer to the attachment!!

Attachments:
Answered by ObnoxiousS
52

Given :-

  • Acceleration of the train = 5 m/s²
  • Time taken = 5 seconds

To Find :-

  • Velocity of the train
  • Distance covered by the train

Formula used :-

\bf \longrightarrow\red { v = u + at} \: (First \:  equation \:  of \:  motion) \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \\ \bf \longrightarrow\pink{s = ut +  \frac{1}{2} a {t}^{2} } \: ( Second \:  equation \:  of  \: motion)

Where,

  • v = final velocity
  • u = initial velocity
  • t = time taken
  • s = distance covered
  • a = acceleration

First of all, we will find velocity of the train,

  \longrightarrow \sf \: v = u + at \\ \longrightarrow \sf \:  0 + 5  \times 5 \:  \:  \\\longrightarrow \sf v = 25 \: m/  {s}

Now, for the distance,

 \sf \longrightarrow \: ut +  \frac{1}{2} a   {t}^{2}  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \\ \sf \longrightarrow \: 0 \times 5 +  \frac{1}{2}  \times 5 \times  {5}^{2}  \\ \sf \longrightarrow \:  \frac{1}{2}  \times  {5}^{3}  =  \frac{1}{2}   \times 125   \:\\ \sf \longrightarrow \frac{1}{ \cancel2}   \times  \cancel{125} = 62.5 \: m \:  \:

\sf\red {\therefore  velocity = 25 m/s}  \:  \:  \: \\ \sf \pink {and  \: distance = 62.5 m}

✯  \: \bf \underline {Important :  - }

There are three equations of motion :-

\bf \longrightarrow\red { v = u + at} \: (First \:  equation \:  of \:  motion) \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \\ \bf \longrightarrow\pink{s = ut +  \frac{1}{2} a {t}^{2} } \: ( Second \:  equation \:  of  \: motion) \\ \bf \longrightarrow\blue {{v}^{2}  =  {u}^{2}  + 2as} \: ( Third  \: equation  \: of  \: motion) \:

 \sf{↦ First \:  equation \:  of  \: motion  \: is  \: also  \: known \:  as \:  " Velocity - time \:  relation" } \:  \:  \:  \:  \:  \:  \\  \sf↦Second \:  equation  \: of  \: motion \:  is \:  also  \: known \:  as  \: " \: position - time \:  relation".  \\  \sf↦Third \:  equation \:  of \:  motion \:  is  \: also  \: known \:  as  \: "velocity - position  \: relation"

Similar questions