Question

A particle is moving in a straight line, where its position (in meter) is a function of time $t$ (in seconds) given by $s=a{{t}^{2}}+bt+6\ge 0$. If it is known that the particle comes to rest after $4$ seconds at a distance of $16$ meters from the starting position ($t=0$), then the retardation in its motion is
$1)\text{ }-1m/{{s}^{2}}$
$2)\text{ }\left( 5/4 \right)m/{{s}^{2}}$
$3)\text{ }\left( 1/2 \right)m/{{s}^{2}}$
$4)\text{ }\left( -5/4 \right)m/{{s}^{2}}$

Answer 1

In this question we have been given with the equation of the distance of a particle. We will use the equation to find the equation for velocity is given by the formula $v=\dfrac{d}{dt}s$ and also the equation for acceleration which is given by $a=\dfrac{d}{dt}v$. We will then substitute the data given in the question in the expression and find the retardation and get the required solution.

Complete step by step answer:
Consider the distance travelled be $s$, the velocity be $v$ and the acceleration as $x$.
We have the equation for the distance given to us as:
$\Rightarrow s=a{{t}^{2}}+bt+6\to \left( 1 \right)$
Now we know that velocity is given as $v=\dfrac{d}{dt}s$ therefore, on substituting, we get:
$\Rightarrow v=\dfrac{d}{dt}\left( a{{t}^{2}}+bt+6 \right)$
On differentiating, we get:
$\Rightarrow v=2at+b\to \left( 2 \right)$
Now we know that acceleration is given as $x=\dfrac{d}{dt}v$ therefore, on substituting, we get:
$\Rightarrow x=\dfrac{d}{dt}\left( 2at+b \right)$
On differentiating, we get:
$\Rightarrow x=2a\to \left( 3 \right)$
Now we know from the question that after $4$ seconds, the velocity is $0$ and the distance travelled is $16m$.
On substituting the values in equation $\left( 2 \right)$, we get:
$\Rightarrow 0=2a\times 4+b$
On simplifying, we get:
$\Rightarrow 0=8a+b$
On rearranging, we get:
$\Rightarrow b=-8a\to \left( 4 \right)$
Similarly, on substituting the values in equation $\left( 1 \right)$, we get:
$\Rightarrow 16=a\times {{\left( 4 \right)}^{2}}+4b+6$
On simplifying, we get:
$\Rightarrow 16=16a+4b+6$
Now from equation $\left( 4 \right)$, we get $b=-8a$ therefore, on substituting, we get:
$\Rightarrow 16=16a+4\left( -8a \right)+6$
On multiplying, we get:
$\Rightarrow 16=16a-32a+6$
On simplifying, we get:
$\Rightarrow a=-\dfrac{5}{8}$
Now from equation $\left( 3 \right)$, we have acceleration as:
$\Rightarrow x=2a$
On substituting the value of $a$, we get:
$\Rightarrow x=2\left( -\dfrac{5}{8} \right)$
On simplifying, we get:
$\Rightarrow x=-\dfrac{5}{4}$, which is the required value of acceleration.
Now since negative acceleration is called retardation, we have the solution as $-5/4m/{{s}^{2}}$.

So, the correct answer is “Option 4”.

Note: It is to be remembered that acceleration and retardation are the same concept just with different signs. Acceleration is when the speed of a moving object increases with the increase in time and retardation is when the speed of a moving object decreases with the increase in time. It is to be noted that retardation is also called deceleration.