Question

starting from the rest train moves with uniform acceleration of 1.5m per second square for 3 minute calculate the total distance covered by the train
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Answer 1

GiveN :

• Initial velocity (u) = 0 m/s
• Acceleration (a) = 1.5 m/s²
• Time (t) = 3 min = 180 sec

To FinD :

• Distance covered by the train

SolutioN :

Here we are given initial velocity, acceleration and time. We can find out the value of Distance traveled by using Kinematics Equations.

$\underbrace{\sf{Distance \: Traveled \: by \: Train}}$

Use 2nd equation of Kinematics

$\implies \sf{S \: = \: ut \: + \: \dfrac{1}{2} at^2} \\ \\ \implies \sf{S \: = \: 0 \: \times \: 180 \: + \: \dfrac{1}{2} \: \times \: 1.5 \: \times \: (180)^2} \\ \\ \implies \sf{S \: = \: \dfrac{1.5 \: \times \: 32400}{2}} \\ \\ \implies \sf{S \: = \: 1.5 \: \times \: 16200 } \\ \\ \implies \sf{S \: = \: 24300} \\ \\ \underline{\underline{\sf{Distance \: Travelled \: by \: train \: is \: 24300 \: m \: or \: 24.3 \: km}}}$

Answer 2

Answer:

24.3 kilometres

Explanation:

Given:

Initial velocity (u) = 0 m/s

Acceleration (a) = 1.5 m/s²

Time = 3 minutes

1 minute = 60 seconds

3 minutes = 3×60 seconds = 180 seconds (t)

To find:

Distance covered by train

Using the second equation of motion, which says:

S=ut+$\frac{1}{2} at^{2}$

Substituting the above values, we get:

S=0×180+$\frac{1}{2} \times 1.5 \times 180 \times 180$

S=$\frac{1}{2} \times 1.5 \times 180 \times 180$

S= 24300 metres

S= 24.3 kilometres

The train will cover a distance which is equal to 24.3 kilometres