# Kinematics Example | Physics for Engineers Lesson 4 Part 2

A car accelerates uniformly from rest. Its acceleration is given by a(t) = 4t m/s², where t is the time in seconds. Find the car’s velocity and position as functions of time.

Step 1: Find the velocity function by integrating the acceleration function with respect to time (t).

v(t) = ∫ a(t)

dt = ∫ 4t dt

Integrating, we get:

v(t) = 2t² + C₁

Since the car starts from rest, its initial velocity v(0) = 0. Therefore, C₁ = 0.

v(t) = 2t²

Step 2: Find the position function by integrating the velocity function with respect to time (t).

x(t) = ∫ v(t) dt = ∫ 2t² dt

Integrating, we get:

x(t) = (2/3)t³ + C₂

Since the car starts at the origin, its initial position x(0) = 0. Therefore, C₂ = 0.

x(t) = (2/3)t³

Now we have the velocity and position functions of the car as a function of time:

v(t) = 2t² (m/s)
x(t) = (2/3)t³ (m)

These functions describe the car’s velocity and position at any given time t during its uniform acceleration.

## Activity

A car accelerates uniformly from rest, following the acceleration function a(t) = 5t m/s², where t is the time in seconds. Find the car’s velocity when its position is 10 meters from the starting point.