Question

How long does it take a car to cross a $30$ meter wide intersection after the light turns green, if the car accelerates from at a constant $2.00m/{{s}^{2}}?$

Answer 1

We are given that the acceleration is constant. We know that the displacement is also fixed. We will use one of the equations of motion $S=ut+\dfrac{1}{2}a{{t}^{2}}$ to find the time. We will apply the values in the equation and make some rearrangements.

Complete step-by-step solution:
Let us consider the given problem.
We are asked to find the time a car takes to cross a $30$ meter wide intersection after the light turns, if the car accelerates from at a constant $2.00m/{{s}^{2}}.$
We can see that the displacement and the acceleration are fixed.
The displacement is given by $S=30m$ and the acceleration is given by $a=2.00m/{{s}^{2}}.$
We will use one of the equations of motion to find the time. So, the equation we are using here is $S=ut+\dfrac{1}{2}a{{t}^{2}}.$
In this equation, $S$ is the displacement, $a$ is the acceleration and $u$ is the initial velocity.
We know that at the starting position, the velocity is zero. Therefore, the initial velocity of the car is $u=0m/s.$
Let us apply the values in the equation to get $30=0\times t+\dfrac{1}{2}\times 2\times {{t}^{2}}.$
This will give us $30=\dfrac{1}{2}\times 2\times {{t}^{2}}.$
And so, when we cancel the common term $2,$ we will get $30={{t}^{2}}.$
We will get $t=\sqrt{30}.$
We will get the time as $t=5.477.$
Hence the time the car will take to cross a $30$ meter wide intersection after the light turns green, if the car accelerates from at a constant $2.00m/{{s}^{2}}$ is $5.5s.$

Note: The equations of the motion are: $v=u+at, S=ut+\dfrac{1}{2}a{{t}^{2}}, {{v}^{2}}-{{u}^{2}}=2aS.$ In these equations, $S$ is the displacement, $u$ is the initial velocity, $v$ is the final velocity, $t$ is the time taken and $a$ is the acceleration.