Variable Acceleration : A Level Maths
Variable Acceleration is a vital topic in A-Level Maths. Unlike constant acceleration, it allows acceleration to change over time, position, or velocity. This makes it more realistic for many physical situations.
- What you need to know: You need calculus (both differentiation and integration) to describe how displacement, velocity, and acceleration are related.
Introduction to Variable Acceleration
Until now, we have primarily used SUVAT EQUATIONS, which assume a constant acceleration. However, in real-life situations, acceleration is rarely constant. For example:
- A falling stone faces air resistance, causing acceleration to change.
- A car experiences varying forces like friction and engine power.
In such cases, we need calculus-based methods—differentiation and integration—to model variable acceleration accurately.
Using Differentiation: From Displacement to Acceleration
From Displacement to Velocity
- Velocity measures how quickly displacement changes with respect to time. For a finite time interval:
- As Δt becomes very small, we use the derivative form:
- Hence, to find velocity, differentiate displacement with respect to time.
Example:
If s(t) = t3 , then
From Velocity to Acceleration
- Acceleration measures how quickly velocity changes with respect to time. For a finite time interval:
- As Δt becomes very small, we use:
- Hence, to find acceleration, differentiate velocity, or equivalently take the second derivative of displacement.
Example:
If v(t) = 2t3 + 4t, then
Using Integration: From Acceleration to Displacement
From Acceleration to Velocity
- Since acceleration is the rate of change of velocity, we can find velocity by integrating acceleration with respect to time:
Example:
If a(t) = 6t, then
- Here, C is the constant of integration, which represents the initial velocity.
From Velocity to Displacement
- Similarly, displacement can be found by integrating velocity with respect to time:
Example:
If v(t) = 3t2 +4, then
- Here, C is again a constant that signifies the initial displacement.
Equations of Motion Under Variable Acceleration
Past Paper Solved Example
Past Paper Solved Example
Problem:
At time t seconds, where t > 0, a particle P has velocity v m/s−1 where
(a) Find the speed of P at time t = 2 seconds.
(b) Find an expression, in terms of t, i, and j, for the acceleration of P at time t seconds, where t > 0.
At time t = 4 seconds, the position vector of P is (i − 4j)m.
(c) Find the position vector of P at time t = 1 second.
(Source: Edexcel A-Level Mechanics, June 2022)
Solution:
Part (a): Find the Speed of P at t = 2 Seconds
Step #1: Express the given velocity vector:
Step #2: Calculate the magnitude of v to find the speed, which is given by the formula:
Step #3: Substitute t = 2 into the equation:
Part (b): Expression for Acceleration of P at Time t
Step #1: Start with the given velocity vector:
Step #2: Differentiate each component of v with respect to t to find the acceleration vector a:
Step #3: Combine the components to form the acceleration vector:
Part (c): Finding the Position Vector of P at t = 1 Second
- Given that the position vector at t = 4 seconds is (i − 4j)m, we first need to integrate the velocity function to find the general form of the position vector and then use the given information to solve for any constants.
Step #1: Express the velocity function:
Step #2: Integrate the velocity vector to find the position vector s(t):
Where C is a vector constant.
Step #3: Use the given position at t = 4 to find C:
Solve for C
Step #4: Substitute C back into the position vector equation:
Step #5: Evaluate s(t) at t = 1:
Therefore, the position vector of P at t = 1 second is (-62i + 24j) meters
Past Paper Solved Example
Problem:
[In this question position vectors are given relative to a fixed origin O]
At time t seconds, where t ≥ 0, a particle, P, moves so that its velocity v m/s−1 is given by:
When t = 0, the position vector of P is (−20i + 20j)m.
(a) Find the acceleration of P when t = 4.
(b) Find the position vector of P when t = 4.
(Source: Edexcel A-Level Mechanics, June 2019)
Solution:
Part (a): Finding the Acceleration at t = 4 Seconds
Step #1: Start with the velocity vector:
Step #2: Differentiate the velocity vector to find acceleration:
Step #3: Evaluate a at t = 4:
Part (b): Finding the Position Vector at t = 4 Seconds
Step #1: Integrate the velocity vector to find the position vector:
Step #2: Use the initial condition s(0) = −20i + 20j to solve for C:
Step #3: Substitute t = 4 into the position function:
Video Tutorial on Variable Acceleration
- Watch this Video Tutorial as we explain step by step to Find Variable Acceleration
Practice Questions and Answers on Variable Acceleration
Question 1: A particle moves along a straight line with velocity given by v = 3t2 − 4t + 2 m/s. Find the acceleration at t = 3 seconds.
Question 2: The displacement of a particle is given by s=2t3 − 5t2 + 4t, where s is in meters and t is in seconds. Find the velocity and acceleration at t = 2 seconds.
Question 3: A particle moves such that its acceleration is given by a = 6t − 4. If its initial velocity is 3 m/s when t = 0, find the velocity function v(t).
Question 4: A particle moves with acceleration a = 12t2 − 6t. Given that its velocity is 5 m/s when t = 1, find the velocity when t = 3.
Question 5: The velocity of a particle moving along a straight line is given by v = 4t3 − 3t2 + 2. Determine the time when the acceleration is zero.
Question 6: A body moves such that its acceleration is given by a = 3t + 2. If the initial velocity is 4 m/s and initial displacement is 10 m when t = 0, find an expression for displacement s(t).
Question 7: A particle moves along a straight path, and its acceleration is given by a = 5t2 − 3t + 1. Find the change in velocity between t = 2 and t = 4 seconds.
Question 8: The velocity of a moving particle is given by v = t2 − 4t + 6. Determine the time interval when the particle is moving in the positive direction.
Question 9: The position of a particle moving along a straight line is given by s = t4 − 6t3 + 9t2 . Find the time at which the acceleration is equal to zero.
Question 10: A particle moves such that its velocity is given by v = 3t2 − 2t. Find the total distance traveled in the first 5 seconds.
Solutions
Question 1:
Solution:
Step #1: Differentiate velocity to find acceleration.
a = d(v)/dt = d(3t² – 4t + 2)/dt
= 6t – 4
Step #2: Substitute t = 3 into the acceleration equation.
a = 6(3) – 4 = 18 – 4 = 14
Final Answer: 14 m/s²
Question 2:
Solution:
Step #1: Differentiate displacement to get velocity.
v = d(s)/dt = d(2t³ – 5t² + 4t)/dt
= 6t² – 10t + 4
Step #2: Evaluate velocity at t = 2.
v(2) = 6(2)² – 10(2) + 4
= 24 – 20 + 4 = 8
Step #3: Differentiate velocity to get acceleration.
a = d(v)/dt = d(6t² – 10t + 4)/dt
= 12t – 10
Step #4: Evaluate acceleration at t = 2.
a(2) = 12(2) – 10 = 24 – 10 = 14
Final Answer: Velocity = 8 m/s, Acceleration = 14 m/s²
Question 3:
Solution:
Step #1: Integrate acceleration to get velocity.
∫(6t – 4) dt = 3t² – 4t + C
Step #2: Solve for C using v(0) = 3.
3 = 3(0)² – 4(0) + C
C = 3
Final Answer: v = 3t² – 4t + 3
Question 4:
Solution:
Step #1: Integrate acceleration to get velocity.
∫(12t² – 6t) dt = 4t³ – 3t² + C
Step #2: Solve for C using v(1) = 5.
5 = 4(1)³ – 3(1)² + C
5 = 4 – 3 + C
C = 4
Step #3: Evaluate velocity at t = 3.
v(3) = 4(3)³ – 3(3)² + 4
= 4(27) – 3(9) + 4
= 108 – 27 + 4 = 85
Final Answer: v = 85 m/s
Question 5:
Solution:
Step #1: Differentiate velocity to get acceleration.
a = d(v)/dt = d(4t³ – 3t² + 2)/dt
= 12t² – 6t
Step #2: Solve for t when a = 0.
12t² – 6t = 0
6t(2t – 1) = 0
Step 3: Solve for t.
t = 0 or t = 1/2
Final Answer: t = 0 or t = 1/2 seconds
Question 6:
Solution:
Step #1: Integrate acceleration to get velocity.
∫(3t + 2) dt = (3/2)t² + 2t + C
Step #2: Solve for C using v(0) = 4.
4 = 0 + 0 + C
C = 4
Step #3: Integrate velocity to get displacement.
∫((3/2)t² + 2t + 4) dt = (1/2)t³ + t² + 4t + D
Step #4: Solve for D using s(0) = 10.
10 = 0 + 0 + 0 + D
D = 10
Final Answer: s = (1/2)t³ + t² + 4t + 10
Question 7:
Solution:
Step #1: Integrate acceleration to get velocity.
∫(5t² – 3t + 1) dt = (5/3)t³ – (3/2)t² + t + C
Step #2: Find v(4) – v(2).
Δv = v(4) – v(2)
= [(5/3)(4)³ – (3/2)(4)² + 4] – [(5/3)(2)³ – (3/2)(2)² + 2]
Final Answer: Change in velocity = 78 m/s
Question 8:
Solution:
Step #1: Find roots of velocity equation.
t² – 4t + 6 = 0
- Discriminant = (-4)² – 4(1)(6) = 16 – 24 = -8 (negative)
- Since there are no real roots, velocity is always positive.
Final Answer: Velocity is always positive for all t.
Question 9:
Solution:
Step #1: Differentiate displacement to get velocity.
v = d(s)/dt = d(t⁴ – 6t³ + 9t²)/dt
= 4t³ – 18t² + 18t
Step #2: Differentiate velocity to get acceleration.
a = d(v)/dt = d(4t³ – 18t² + 18t)/dt
= 12t² – 36t + 18
Step #3: Solve for t when acceleration is zero.
12t² – 36t + 18 = 0
Divide by 6:
2t² – 6t + 3 = 0
Factor or use quadratic formula:
t = 1.5
Final Answer: t = 1.5 seconds
Question 10:
Solution:
Step #1: Integrate velocity to get displacement.
s = ∫(3t² – 2t) dt
= (t³ – t²) + C
Step #2: Evaluate displacement at t = 5 and t = 0.
s(5) – s(0) = (5³ – 5²) – (0³ – 0²)
= (125 – 25) – 0
= 100
Final Answer: Total distance traveled = 100 meters