Question

A train starting from rest moves with a uniform acceleration of 0-2m/square for 5 minutes. calculate the speed acuired and distance travelled in this time
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Answer 1

Given:-

• Initial velocity =0m/s

• Acceleration =2m/s²

• Time taken =5min =5×60=300s

To Find:-

• Final velocity

• Distance travelled train

Solution :-

By using 1st equation of motion

➦ v=u+at

➭ v = 0+ 2× 300

➭ v=600m/s

∴The final velocity of train is 600m/s

and Now,

By using 2nd equation of motion

➦ s=ut +1/2at²

➭ s = 0×300 +1/2 ×2 × 300×300

➭ s =90000 m

∴The distance covered by train is 90km.

Answer 2

$\rule{200}2$

$\huge{\mathtt{QUESTION}}$

➡ A train starting from rest moves with a uniform acceleration of 0-2m/square for 5 minutes. Calculate the speed aquired and distance travelled in this time .

$\rule{200}2$

$\huge{\mathtt{SOLUTION}}$

$\large{\textsf{\red{Given\: :}}}$

$\sf Initial \:velocity\: = \:0\: ( as\: the \:train\: was \:at \:rest )$

$\sf Acceleration\: = \:2\: {m}^{2}$

$\sf Time\:=\: 5\:min. \:=\: 5 \times 60 \:=\: 300\:sec$

$\large{\textsf{\red{To\:Find\: :}}}$

↪Speed

↪Distance travelled

☯ Now , from first equation of motion →

↪ v = u + at

↪ v = 0 + 2(300)

↪ v = 600 m/s

$\:\:\:\:\:\:\large{\textsf{\red{SPEED\:=\:600\:m/s}}}$

☯ Now , from second equation of motion

↪$\sf {v}^{2} - {u}^{2} = 2as$

↪$\sf {600}^{2} - 0 = 2(2)s$

↪$\sf s = \dfrac{600 \times 600}{2 \times 2}$

↪$\sf s = \dfrac{360000}{4}$

↪$\sf s = 90000$

$\:\:\:\:\:\:\large{\textsf{\red{DISTANCE\:=\:90000\:m\:90\:km}}}$

$\rule{200}2$