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?
PLS HELP ME FAST...
Share
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?
PLS HELP ME FAST...
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Gɪᴠᴇɴ :-
➻ ᴛʀᴀɪɴ's ➘
➻ ᴍᴀɴ's ➘
Tᴏ Fɪɴᴅ :-
sᴏʟᴜᴛɪᴏɴ :-
By using 2nd equation of motion, we get
Put the given values of train in this equation
➦ 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 :-
Here,
↠ a = acceleration
↠ v = final velocity
↠ u = initial velocity
↠ s = distance travelled
↠ t = time taken