A body starts from rest, the ratio of distances travelled by the body ...
The velocity after 2 sec = 2a and velocity after 3 sec = 3a
For some constant acceleration a,
Now distance travelled in third second, s3 = 2a + ½ a
= 5/2 a
Similarly distance travelled in fourth second, s4= 3a + ½ a
= 7/2 a
Hence the reqiuired ratio is 5/7
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A body starts from rest, the ratio of distances travelled by the body ...
In question its asked about ratio of distance travelled in 3rd second & distance travelled in 4th second..
In this case; we have the formula;
Sn=u+a(n-1/2)..
S3=0+a(3-1/2)..
S3=a(6-1/2)..
S3=a(5/2)..
S4=u+a(4-1/2)..
S4=0+a(4-1/2)..
S4=a(7/2)..
S3/S4=a(5/2)/a(7/2)..
S3/S4=5/7..
Hence;option B is true...
A body starts from rest, the ratio of distances travelled by the body ...
Solution:
Given, the body starts from rest.
Let, s1 and s2 be the distances travelled by the body in the 3rd and 4th seconds respectively.
Using the formula, s=ut+1/2at^2, where u= initial velocity=0, a= acceleration=9.8 m/s^2 (constant)
We get, s1=1/2×9.8×(3)^2=44.1m and s2=1/2×9.8×(4)^2=78.4m
Therefore, the ratio of distances travelled by the body during 3rd and 4th seconds is
s2/s1=78.4/44.1=5/7
Hence, option B is the correct answer.
Answer: The correct option is B) 5/7.
Explanation:
Given, the body starts from rest.
Let, s1 and s2 be the distances travelled by the body in the 3rd and 4th seconds respectively.
Using the formula, s=ut+1/2at^2, where u= initial velocity=0, a= acceleration=9.8 m/s^2 (constant)
We get, s1=1/2×9.8×(3)^2=44.1m and s2=1/2×9.8×(4)^2=78.4m
Therefore, the ratio of distances travelled by the body during 3rd and 4th seconds is
s2/s1=78.4/44.1=5/7
Hence, option B is the correct answer.