The ability to accurately model and predict motion is a cornerstone of physics. The process of using calculus to solve 1D kinematics provides a robust framework for analyzing the movement of objects under variable acceleration. This approach enables us to determine how the position and velocity of an object change over time. Understanding these principles opens the door to solving complex problems.

---

---

Mastering 1D Kinematics with Variable Acceleration

Understanding motion in one dimension with variable acceleration requires calculus. By integrating acceleration, we derive velocity; by integrating velocity, we obtain position. Applying initial conditions anchors the functions to physical reality. This approach allows us to precisely calculate where and when a particle is, even under changing acceleration.

Step 1: Integrating Acceleration to Get Velocity

Given:

### a(t) = 0.6t^2 – 3t + 5 \;\; \text{(m/s²)} ###

Integrate with respect to time:

### v(t) = \int a(t)\,dt = 0.2t^3 – 1.5t^2 + 5t + C. ###

Using the condition ##v(0)=-1## m/s:

### -1 = 0 + C \;\;\Rightarrow\;\; C=-1. ###

Thus:

### v(t) = 0.2t^3 – 1.5t^2 + 5t – 1. ###

Step 2: Integrating Velocity to Get Position

Now integrate velocity:

### x(t) = \int v(t)\,dt = 0.05t^4 – 0.5t^3 + 2.5t^2 – t + D. ###

Using the condition ##x(0)=2## m:

### 2 = D \;\;\Rightarrow\;\; D=2. ###

Thus:

### x(t) = 0.05t^4 – 0.5t^3 + 2.5t^2 – t + 2. ###

Step 3: Finding the Time When x(t)=100 m

Solve:

### 0.05t^4 – 0.5t^3 + 2.5t^2 – t + 2 = 100. ###

Simplified:

### 0.05t^4 – 0.5t^3 + 2.5t^2 – t – 98 = 0. ###

This quartic is solved numerically. The first positive root:

### t \approx 6.61\;\text{s}. ###

Step 4: Finding the Speed at that Instant

Substitute into velocity:

### v(6.61) = 0.2(6.61)^3 – 1.5(6.61)^2 + 5(6.61) – 1 \approx 43.1\;\text{m/s}. ###

So, when the particle reaches 100 m, its speed is about **43.1 m/s**.

Key Takeaways

• Use calculus: integrate acceleration → velocity → position.
• Apply initial conditions to find constants of integration.
• Position equations may require numerical root-finding for exact times.
• Velocity at that instant gives the particle’s speed.

Summary Table of Main Problem

Variable	Expression	Value at t ≈ 6.61 s
Acceleration, a(t)	### 0.6t^2 – 3t + 5 ###	21.2 m/s²
Velocity, v(t)	### 0.2t^3 – 1.5t^2 + 5t – 1 ###	43.1 m/s
Position, x(t)	### 0.05t^4 – 0.5t^3 + 2.5t^2 – t + 2 ###	100 m

Similar Problems (Quick Solutions)

Problem 1: ##a(t)=2t+1,\; v(0)=3,\; x(0)=0.##

Answer: ### v(t)=t^2+t+3,\;\; x(t)=\tfrac{1}{3}t^3+\tfrac{1}{2}t^2+3t. ###

Problem 2: ##a(t)=\cos t,\; v(0)=0,\; x(0)=1.##

Answer: ### v(t)=\sin t,\;\; x(t)=-\cos t + t + 2. ###

Problem 3: ##a(t)=e^{-t},\; v(0)=1,\; x(0)=0.##

Answer: ### v(t)=-e^{-t}+2,\;\; x(t)=2 – e^{-t} – t. ###

Problem 4: ##a(t)=4t-2,\; v(1)=2,\; x(1)=1.##

Answer: ### x(t)=\tfrac{2}{3}t^3 – t^2 + \tfrac{4}{3}t. ###

Problem 5: ##a(t)=6t^2,\; v(0)=-2,\; x(0)=3.##

Answer: ### x(t)=0.5t^4 – 2t + 3. ###

Below are quick integrations of acceleration functions to obtain velocity and position, applying given initial conditions. These examples highlight how calculus simplifies 1D kinematics with variable acceleration.

Problem 1

Given: ##a(t)=2t+1,\; v(0)=3,\; x(0)=0##

Integrate acceleration: ### v(t)=t^2+t+3. ###

Integrate velocity: ### x(t)=\tfrac{1}{3}t^3+\tfrac{1}{2}t^2+3t. ###

Problem 2

Given: ##a(t)=\cos t,\; v(0)=0,\; x(0)=1##

Integrate acceleration: ### v(t)=\sin t. ###

Integrate velocity: ### x(t)=-\cos t + t + 2. ###

Problem 3

Given: ##a(t)=e^{-t},\; v(0)=1,\; x(0)=0##

Integrate acceleration: ### v(t)=-e^{-t}+2. ###

Integrate velocity: ### x(t)=2 – e^{-t} – t. ###

Problem 4

Given: ##a(t)=4t-2,\; v(1)=2,\; x(1)=1##

Integrate acceleration with condition: ### v(t)=2t^2-2t. ###

Integrate velocity with condition: ### x(t)=\tfrac{2}{3}t^3 – t^2 + \tfrac{4}{3}t. ###

Problem 5

Given: ##a(t)=6t^2,\; v(0)=-2,\; x(0)=3##

Integrate acceleration: ### v(t)=2t^3 – 2. ###

Integrate velocity: ### x(t)=0.5t^4 – 2t + 3. ###

Summary Table

Problem	Acceleration a(t)	Velocity v(t)	Position x(t)
1	##2t+1##	##t^2+t+3##	##\tfrac{1}{3}t^3+\tfrac{1}{2}t^2+3t##
2	##\cos t##	##\sin t##	##- \cos t + t + 2##
3	##e^{-t}##	##-e^{-t}+2##	##2 – e^{-t} – t##
4	##4t-2##	##2t^2-2t##	##\tfrac{2}{3}t^3 – t^2 + \tfrac{4}{3}t##
5	##6t^2##	##2t^3-2##	##0.5t^4 – 2t + 3##

---

---