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A body travels 2m in the 2nd second and 6m in the next 4 seconds. What will be the distance travelled in 9th second.
Solution: As per the given data we have distance covered in 2nd sec = 2m and distance travelled is 6m in 4 sec. So, at the end of 9th sec the distance would be 9m.Thanks & Regards,Vasantha Sivaraj, askIITians faculty.
Solution:
As per the given data we have distance covered in 2nd sec = 2m and distance travelled is 6m in 4 sec.
So, at the end of 9th sec the distance would be 9m.
Thanks & Regards,
Vasantha Sivaraj,
askIITians faculty.
Let's consider initial velocity is u and acceleration is a and 8th sec velocity is v..we will this equation S=ut+0.5at2 as shown below2 = u*2 +0.5 *a*2*2 -----1 8 = u*6 + 0.5*a*6*6 ------2now 3x(2 = u*2 +0.5 *a*2*2 ) ----3now we will subtract from equation 3 to equation 2we will get a = 1/6 m/s^2Putting above value of a in eq 2 we get8 = u*6+0.5*1/6*6*6u = 5/6 m/s^2v = u+atNow, v = 5/6+ 1/6*9 So, v=7/3 m/sThanks & Regards,Nirmal SinghAskiitians Faculty
Let's consider initial velocity is u and acceleration is a and 8th sec velocity is v..
we will this equation S=ut+0.5at2 as shown below
2 = u*2 +0.5 *a*2*2 -----1 8 = u*6 + 0.5*a*6*6 ------2
now 3x(2 = u*2 +0.5 *a*2*2 ) ----3
now we will subtract from equation 3 to equation 2
we will get a = 1/6 m/s^2
Putting above value of a in eq 2 we get
8 = u*6+0.5*1/6*6*6
u = 5/6 m/s^2
v = u+at
Now, v = 5/6+ 1/6*9 So, v=7/3 m/s
Nirmal Singh
Askiitians Faculty
1 ) 2s it covers 2m .2) 6s (next 4 sec means 2+4) it covers 8m (2m+6m) ...compare the ratios ...6s/2s : 8m/2s we get, 3:4 ...now to find dist travelled in 9s x/4=9/3 x=12m
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