if a particle starting with certain initial velocity and uniform accelration covers a distance 12m in first 3 seconds and a distance of 30m in next 3 seconds.then find its inital velocity
(vi+v3)/2 = 12/3 = 4 and (v3+v6)/2 = 30/3=10
so
v3 =8-vi and v3 = 20-v6
and
v6=vi+6a
so
8-vi = 20 -vi -6a
or
6a = 12
a = 2
then
vi + 3(2) = v3 and vi+v3 = 8
so
vi+6 = 8-vi
2 vi = 2
vi = 1
I am not so sure about the answer but I think :
D1=12m ,T1=3secs ,D2=30m ,T2=3secs
Initial velocity =
V=d1/t1=12m/3secs=4m/s
Please bear with me for I just showed my opinion, any mistake can be corrected .Thanks
To find the initial velocity of the particle, we can use the equations of motion.
First, let's define the variables:
- u: initial velocity (what we want to find)
- a: uniform acceleration of the particle
- t1: time taken to cover the first distance of 12m
- t2: time taken to cover the next distance of 30m
Now, let's use the equations of motion:
1. Distance covered in terms of initial velocity, acceleration, and time:
d = ut + (1/2)at^2
For the first 3 seconds:
12 = u * 3 + (1/2) * a * (3^2)
12 = 3u + (1/2) * 9a [equation 1]
For the next 3 seconds:
30 = u * 6 + (1/2) * a * (6^2)
30 = 6u + (1/2) * 36a [equation 2]
2. Now, let's solve the above two equations to find the values of u and a.
From equation 1, we get:
12 = 3u + (1/2) * 9a
24 = 6u + 9a
Subtracting this from equation 2, we eliminate 'u':
30 - 24 = 6u + (1/2) * 36a - (6u + 9a)
6 = (1/2) * 27a
12 = 27a
a = 12/27
a = 4/9
3. Now, substitute the value of 'a' in equation 1:
12 = 3u + (1/2) * 9a
12 = 3u + (1/2) * 9 * (4/9)
12 = 3u + (1/2) * 4
12 = 3u + 2
10 = 3u
u = 10/3
Therefore, the initial velocity (u) of the particle is 10/3 m/s.