# David is driving a steady 25.0 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2.00 m/s^2 at the instant when David passes. How far does Tina drive before passing David, and what is her speed as she passes him?

This question aims to find the displacement and speed of the car.

Distance regards the total motion of an object without having any direction. It can be defined as the amount of surface an object has obscured regardless of its initial or endpoint. It is the numerical estimation of how far away an object is from a specific point. Distance refers to physical length or an estimation based on some factors. Furthermore, the factors under consideration for the distance to be calculated include speed and the time to cover a specific distance. Displacement is referred to as the variation in the object’s position. It is a vector quantity having a magnitude and direction. It is symbolized by an arrow that points from the initial point to the ending point. For instance, the movement of an object from one point to another result in a change in its position, and this change is said to be displacement.

Speed and velocity describe the slow or fast movement of an object. We frequently meet situations where we must determine which of two objects is traveling much faster. If they are consequently traveling in the same direction and on the same path, it is easy to say which object is going faster. Moreover, determining the fastest object is challenging if the motions of two are in opposite directions.

The formula for the displacement of an object is given by:

$s(t)=ut+\dfrac{1}{2}at^2$

Initially, Tina’s car is at rest, therefore:

$(25\,m/s)t=0+\dfrac{1}{2}(2.00\,m/s^2)t^2$

$t=25\,s$

Now, use the same formula to find the displacement as:

$s(t)=0+\dfrac{1}{2}(2.00\,m/s^2)(25\,s)^2$

$s(t)=625\,m$

Tina’s speed when she passes David can be calculated as:

$v=at$

$v=(2.00\,m/s^2)(25\,s)$

$v=50\,m/s$

## Example 1

Assume that a cat runs from one point on the road to the other point at the end of the road. The length of the road is $75\,m$ in total. Furthermore, it takes $23\,s$ to cross the end of the road. Determine the speed of the cat.

### Solution

Let $s$ be the speed, $d=75\,m$ be the distance and $t=23\,s$ be the time. The formula for the speed is given by:

$s=\dfrac{d}{t}$

Now, substitute the given values as:

$s=\dfrac{75\,m}{23\,s}$

$s=3.26\,m/s$

Hence, the speed of the cat will be $3.26\,m/s$.