Question

As a car passes the point $A$ on a straight road, its speed is $10\text{m/s}$. The car moves with constant acceleration $a{\text{m/s}}^2$ along the road for $T$ seconds until it reaches the point $B$, where its speed is $v\text{m/s}$. The car travels at this speed for a further $10$ seconds, when it reaches the point $C$. From $C$ it travels for a further $T$ seconds with constant acceleration $3a{\text{m/s}}^2$ until it reaches a speed of $20\text{m/s}$ at the point $D$. Choose the correct option(s).
[Given that the distance between $A$ and $D$ is $675\text{m}$]

• A. value of $v$ is $12.5\text{m/s}$
• B. value of $a$ is $\frac{1}{8}{\text{m/s}}^2$
• C. value of $T$ is $10\text{s}$
• D. value of $T$ is $20\text{s}$

Answer 1

Answer: (A), (B), (D)

value of $T$ is $20\text{s}$

Let us try to understand the situation given in the question with the help of graphical representation.

From $A$ to $B$, $u=10\text{m/s}$, acceleration $=a{\text{m/s}}^2$
Using first equation of motion, we can say
$10+aT=v\dots \dots \dots \dots \dots \left(1\right)$
From $C$ to $D$, acceleration $=3a{\text{m/s}}^2$
Using first equation of motion, we can say
$u+a_{2}t=20\text{m/s}$
$\Rightarrow (10+aT)+3aT=20$
$\Rightarrow 10+4aT=20\dots \dots \dots \dots \left(2\right)$
From $\left(1\right)$ and $\left(2\right)$,
$v=12.5\text{m/s}.........\left(3\right)$
Distance between $A$ and $D$ $=675\text{m}$ = Area under the curve
= Area under line $AB$ + Area under line $BC$ + Area under line $CD$
$=\frac{1}{2}T(10+v)+10v+\frac{1}{2}(v+20)T=675.....\left(4\right)$
Using eq. $\left(3\right)$ in eq. $\left(4\right)$, we get
$T=20\text{s}$.
From eq. $\left(1\right)$, $a=\frac{12.5-10}{20}=\frac{1}{8}{\text{m /s}}^2$
Hence, options (a), (b) and (d) are correct.