Solveeit Logo

Question

Question: A particle moving with a uniform acceleration travels \(24m\) and \(64m\)in the first two constant i...

A particle moving with a uniform acceleration travels 24m24m and 64m64min the first two constant intervals of 4sec4\sec . Its initial velocity is
A. 1m/sec1m/\sec
B. 10m/sec10m/\sec
C. 5m/sec5m/\sec
D. 2.5m/sec2.5m/\sec

Explanation

Solution

The equation of motions describes the basic concept of the object’s motion like position, acceleration, and velocity. These equations govern the object’s motion in 1D,2D and 3D1D,2D{\text{ and }}3D. To solve the given problem, consider the equation of the motions.

Formula used:
Formula to find the initial velocity:
v=u+at\Rightarrow v = u + at
Where, vv is the velocity, aais the acceleration, tt is the time, and uu is the initial velocity.

Complete step by step answer:
A particle is moving with uniform acceleration. The particle has two constant intervals of 4sec4\sec at
24m24m and at 64m64m.
We need to calculate the velocity of each interval first. Initial velocity for the particle at 24m24m of constant interval of 4sec4\sec .
v=u+at\Rightarrow v = u + at
Where, vv is the velocity, aa is the acceleration, tt is the time, and uu is the initial velocity.
Substitute the values t=4sect = 4\sec in the formula.
v=u+4a\Rightarrow v = u + 4a
The value for the velocity is the same for the next interval at 64m64m. That is, v=u+4av = u + 4a
Next, calculate the speed. The values for the speed are given.
For first 4sec4\sec the speed is,
s=ut+12at2\Rightarrow s = ut + \dfrac{1}{2}a{t^2}
Substitute the Values,
24=4u+12a×42\Rightarrow 24 = 4u + \dfrac{1}{2}a \times {4^2}
24=4u+1216a\Rightarrow 24 = 4u + \dfrac{1}{2}16a
Simplify the given equation.
24=4u+8a\Rightarrow 24 = 4u + 8a
4+2a=6\Rightarrow 4 + 2a = 6---- 1
For a second 4sec4\sec the speed is,
s=ut+12at2\Rightarrow s = ut + \dfrac{1}{2}a{t^2}
Substitute the Values,
64=(u+4a)4+12a42\Rightarrow 64 = \left( {u + 4a} \right)4 + \dfrac{1}{2}a{4^2}
Simplify the equation,
16=(u+4a)+2a\Rightarrow 16 = \left( {u + 4a} \right) + 2a
4+6a=16\Rightarrow 4 + 6a = 16--- 2
Subtract equation 2 from 1
We get, 4a=104a = 10therefore, a=2.5a = 2.5substitute the value of acceleration in equation 1.
u+2a=6\Rightarrow u + 2a = 6
u+2(2.5)=6\Rightarrow u + 2\left( {2.5} \right) = 6
u=66\Rightarrow u = \dfrac{6}{6}
u=1\Rightarrow u = 1
Therefore, the initial value is 1m/sec1m/\sec .

So, the correct answer is “Option A”.

Note:
The equation of the motion used to solve the problem is called the first equation of motion. There are totally three types of motion. Equations of the motions are used to solve the kinematics problems. We have to identify the parameters given and then we have to choose the appropriate equations. It is also used to solve the calculations of the optical properties.