Question
Question: A body starts from rest and is uniformly accelerated for \(30\sec \).The distance covered in first \...
A body starts from rest and is uniformly accelerated for 30sec.The distance covered in first 10sec is x1 , for next 10sec is x2 and for last 10sec is x3 the then the ratio of x1:x2:x3?
Solution
In this question we will apply the second equation of motion and with the help of it calculate the distance travelled in each of the 10sec period. Then we will divide the three equations which will be obtained and this will give the required ratio. The second equation of motion is:s=ut+21at2.
Where, s is the displacement, u is the initial velocity, t is the time taken and a is the acceleration of the body.
Complete step by step answer:
According to the second equation of motion,
s=ut+21at2
The distance which is travelled by the body in the first 10sec is,
x1=0×10+21×a×102
x1=21×a×100
x1=50a.......(1)
Similarly, the distance which is travelled by the body in the next 10sec is,
x2=s20−s10
x2=(0×20+21×a×202)−50a
On further solving this, we get,
x2=200a−50a
x2=150a......(2)
Similarly, the distance which is travelled by the body in the last 10sec is,
x3=s30−s20
x2=(0×30+21×a×302)−200a
On further solving this, we get,
x2=450a−200a
x2=250a.....(3)
On comparing the three equations,
x1:x2:x3=50a:150a:250a
On simplifying this,
x1:x2:x3=1:3:5
So, the required ratio is x1:x2:x3=1:3:5.
Note: It is important to note that the three equations of motion are only valid for uniformly accelerated motion. These equations are not valid for a non-uniformly accelerated motion. This is because for a non-uniformly accelerated motion, we cannot associate with the inertial frame of reference. And we also know that, for a non-inertial frame of reference Newton's law is not valid. Hence, we can apply Newton's law of motion only when the body is uniformly accelerated.