Question

At a stop light, a truck travelling at $15m{s}^{-1}$ passes a car as it starts from rest. the truck travels at constant velocity and the car accelerates at $3m{s}^{-2}$ . The vehicles move along parallel straight lines. How many seconds will it take for the car to catch up the truck.

• A. $5s$
• B. $10s$
• C. $15s$
• D. $25s$

Answer 1

The correct option is B

$10s$

Step 1: Given data

A truck is traveling at a velocity $v=15m{s}^{-1}$

The car starts from rest and accelerates at $a=3m{s}^{-2}$

Let the time taken by the car to catch up with the truck be $t$.

Step 2: Determine the distance covered by the truck in time $\mathit{t}$

As we know, $\mathit{d}\mathit{i}\mathit{s}\mathit{t}\mathit{a}\mathit{n}\mathit{c}\mathit{e}\mathbf{}\mathbf{=}\mathbf{}\mathit{v}\mathit{e}\mathit{l}\mathit{o}\mathit{c}\mathit{i}\mathit{t}\mathit{y}\mathbf{\times }\mathit{t}\mathit{i}\mathit{m}\mathit{e}$= 15 x t

Therefore, the distance covered by the truck in time $t$ is:

$s=15t$ $\dots \dots \dots \dots \left(1\right)$

Step 3: Determine the distance covered by the car in time $\mathit{t}$

The car should travel the same distance traveled by truck to catch up with the truck in time $t$ .

As we know the second equation of motion is $s=ut+\frac{1}{2}a{t}^2$

On putting the given data we get,

$\Rightarrow s=\frac{1}{2}\times 3t^2$$\dots \dots \dots \dots \left(2\right)$

Step 4: Determine the time taken by the car to catch up with the truck.

By equating both the equations $\left(1\right)$ and $\left(2\right)$ we get:

$\Rightarrow 15t=\frac{1}{2}\times 3t^2$

$\Rightarrow 30t=3t^2$

On further solving the above equation we get:

$\Rightarrow t^2-10t=0$

$\Rightarrow t\left(t-10\right)=0$

Therefore, by equating the above equation we get two values of $t$ as $t=0s$ and $t=10s$

$t=0s$ is not possible because both car and truck are moving. So, the car will take $10s$ to catch up with the truck.

Final Answer:

It will take $\mathbf{10}\mathbf{}\mathit{s}$ for the car to catch up with the truck. Hence, the correct option is option B.