Question

A train leaves a station from rest with a constant acceleration of 2m/s^2. A man who is 4m behind the train, started running with a constant speed of 4m/s. After what time will he catch the train?


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Answer 1

Gɪᴠᴇɴ :-

➻ ᴛʀᴀɪɴ's ➘

• Initial Velocity (u) = 0 m/s. (start from rest)

• Acceleration (a) = 2 m/s²

➻ ᴍᴀɴ's ➘

• Distance (S) from train = 4m

• Speed (s) = 4 m/s

Tᴏ Fɪɴᴅ :-

• Time taken (t) to catch the train

sᴏʟᴜᴛɪᴏɴ :-

By using 2nd equation of motion, we get

$\Large{\boxed{\tt{s=ut+\frac{1}{2}at^2}}}$

Put the given values of train in this equation

$\implies s = 0 \times t + \frac{1}{2} \times 2 \times {t}^{2} \\ \\ \implies \: s = \frac{2 {t}^{2} }{2} \\ \\ \implies \: s = {t}^{2}$

➦ s = t² ---(1)

Now,

➮ Distance travelled=Speed(s) × Time(t)

Distance travelled by man,

➦ S = 4t. ---(2)

From the question it is given that ,

➮ Difference between distance from man and train = 4m

On subtracting (1) from (2) , we get

↬ S - s = 4

↬ 4t - t² = 4

↬ t² - 4t + 4 = 0

↬ (t) ² - 2 × t × 2 + (2)² = 0

↬ (t - 2)² = 0

↬(t - 2) = 0

↬ t = 2 seconds

Hence,

➲ Time taken by man to catch the train is 2 seconds.

Eǫᴜᴀᴛɪᴏɴs Oғ Mᴏᴛɪᴏɴ :-

• v = u + at

• s = ut + 1/2at²

• 2as = v² - u²

Here,

↠ a = acceleration

↠ v = final velocity

↠ u = initial velocity

↠ s = distance travelled

↠ t = time taken