Question

The law of motion in a straight-line is \[x = \dfrac{1}{2}vt\], then what happens to the acceleration?
A. Constant
B. Proportional to t
C. Proportional to v
D. Proportional to s

Answer 1

Hint:Before we proceed into the problem, it is important to know about the motion in a straight line. Motion in a straight line is a linear motion in which the object changes its position with respect to the time. As the name itself says that it is in a particular straight line. Therefore, it has only one dimension. Now we can solve the problem step by step as follows.

Formula Used:
We know that from the question
\[x = \dfrac{1}{2}vt\]
Where, $v$ is the velocity, $t$ is the time taken and $x$ is the distance from the fixed point on the line.

Complete step by step solution:
\[x = \dfrac{1}{2}vt\]
\[\Rightarrow x = \dfrac{1}{2} \times \dfrac{{dx}}{{dt}} \times t\]
Rearranging the above equation we get,
\[\dfrac{{2dt}}{t} = \dfrac{{dx}}{x}\]

Integrating the above equation on both sides
\[2\log t + \log c = \log x\]
\[\Rightarrow \log {t^2} + \log c = \log x\]............(By using the formula \[A\log B = \log {B^A}\])
\[\Rightarrow \log ({t^2}c) = \log x\]...........(By using the formula \[\log A + \log B = \log AB\])
\[\Rightarrow {t^2}c = x\]
\[ \Rightarrow x = {t^2}c\]

Now differentiate the above equation with respect to t then we get,
\[\dfrac{{dx}}{{dt}} = 2ct\]
\[v = 2ct\].........(Rate of change of distance is velocity, \[\dfrac{{dx}}{{dt}} = v\])
Again, differentiate the above equation with respect to t we get,
\[\dfrac{{dv}}{{dt}} = 2c\]
\[a = 2c\]..........(Rate of change of velocity is acceleration, \[\dfrac{{dv}}{{dt}} = a\])
Since 2c is a constant term, the acceleration is said to be constant. Therefore, the accelerations remain constant.

Hence, Option A is the correct answer

Note:In order to find the solution for this question there is an alternate method which is shown below.
\[x = \dfrac{1}{2}vt\]
Rearrange the above equation for v
\[v = \dfrac{{2x}}{t}\]
Differentiate above equation with respect to t, we get
\[\dfrac{{dv}}{{dt}} = 2(\dfrac{{t\dfrac{{dx}}{{dt}} - x}}{{{t^2}}})\]
\[a = 2(\dfrac{{vt - x}}{{{t^2}}})\]
Substitute the value of x in above equation, we get
\[a = 2(\dfrac{{\dfrac{1}{2}vt}}{{{t^2}}})\]
By simplifying the equation, we get,
\[a = vt\]
On differentiating above equation with respect to t, we obtain,
\[\dfrac{{da}}{{dt}} = (\dfrac{{t\dfrac{{dv}}{{dt}} - v \times 1}}{{{t^2}}})\]
\[\Rightarrow \dfrac{{da}}{{dt}} = (\dfrac{{t \times a - v}}{{{t^2}}})\]
\[\Rightarrow \dfrac{{da}}{{dt}} = \dfrac{0}{{{t^2}}}\]
\[ \Rightarrow \dfrac{{da}}{{dt}} = 0\]
Then \[a = \text{constant}\]
Hence, Option A is the correct answer