Question

A train starts from rest and accelerate uniformly at the rate of 5 m/s2 for 5 sec.Calculate the (a) distance travelled by train in 5 sec. (b) velocity of train at the end of 5 sec​

Answer 1

Given that:

• Initial velocity = 0 mps
• Acceleration = 5 mps sq.
• Time = 5 seconds

To calculate:

• Distance
• Final velocity

Solution:

• Distance = 62.5 m
• Final velocity = 25 mps

Using concepts:

• Second equation of motion
• First equation of motion

Using formulas:

• Second equation of motion:

• s = ut + ½ at²

• First equation of motion:

• v = u + at

Where, s denotes distance or displacement or height, u denotes initial velocity, a denotes acceleration, t denotes time taken and v denotes final velocity.

Required solution:

~ Firstly by using second equation of motion let us find out the distance travelled.

»»» s = ut + ½ at²

»»» s = 0(5) + ½ × 5(5)²

»»» s = 0(5) + ½ × 5(25)

»»» s = 0(5) + ½ × 125

»»» s = 0 + ½ × 125

»»» s = ½ × 125

»»» s = 1 × 62.5

»»» s = 62.5 m

»»» Distance = 62.5 m

~ Now by using first equation of motion let us find out the final velocity!

»»» v = u + at

»»» v = 0 + 5(5)

»»» v = 0 + 25

»»» v = 25 mps

»»» Final velocity = 25 mps

Additional information:

$\begin{gathered}\boxed{\begin{array}{c}\\ {\pmb{\sf{Three \: equation \: of \: motion}}} \\ \\ \sf \star \: v \: = u \: + at \\ \\ \sf \star \: s \: = ut + \: \dfrac{1}{2} \: at^2 \\ \\ \sf \star \: v^2 - u^2 \: = 2as\end{array}}\end{gathered}$