# A car starts from rest and with constant acceleration achieves a velocity of 15 m/s when it...

## Question:

A car starts from rest and with constant acceleration achieves a velocity of 15 m/s when it travels a distance of 200 m. What are

(a) the car's acceleration and

(b) the time required to travel the distance?

## The Kinematic Equations:

Kinematics is the branch of physics that deals with an object's displacement (d), velocity (v), and acceleration (a). These three variables of motion are closely related (each is the integral or derivative of another), and knowing some kinematic information about an object's motion usually lets us solve for more information. We can do this using the kinematic equations:

{eq}v_f = v_i + at\\ v_f^2 = v_i^2 + 2ad\\ d = v_i t + \frac{1}{2}at^2\\ v_{avg} = \frac{1}{2}(v_i + v_f) = \dfrac{d}{t} {/eq}

Here, {eq}t {/eq} is a span of time, {eq}v_i {/eq} is the object's initial velocity, {eq}v_f {/eq} is its final velocity, and {eq}v_{avg} {/eq} is its average velocity. We can use any of the kinematic equations to describe any object, as long as it is moving with a constant acceleration.

(a) This is a kinematics problem. We want to find the car's acceleration {eq}a {/eq}, and we are given its initial velocity (the car starts from rest: {eq}v_i = 0 {/eq}), final velocity ({eq}v_f = 15 \ m/s {/eq}), and displacement ({eq}d = 200 \ m {/eq}). The kinematic equation that connects acceleration, initial velocity, final velocity, and displacement is:

{eq}v_f^2 = v_i^2 + 2ad {/eq}

Let's plug in what we know and solve for acceleration.

{eq}15^2 = 0^2 + 2a(200)\\ 225 = 400a\\ 0.563 \ m/s^2 = a {/eq}

(b) Now, we want to use kinematics to solve for the time that passes as the car travels this distance. Let's consider this formula:

{eq}v_f = v_i + at {/eq}

Now that we have the car's acceleration, we can use this equation to solve for time. Let's plug in what we know.

{eq}15 = 0 + (0.563)t\\ 15 = 0.563t\\ 26.7 \ s = t {/eq} 