Variable Acceleration Using Calculus

Using differentiation and integration when acceleration is not constant

SPEC 7.4 • KINEMATICS

Why Calculus?

Important

SUVAT equations only work when acceleration is constant. When acceleration varies with time, we must use calculus.

In real-world scenarios, acceleration often changes:

Air resistance – increases as speed increases

Springs – force depends on extension

Braking – force may vary with speed

Rocket propulsion – thrust changes as fuel depletes

Key Concept

When displacement, velocity, or acceleration is given as a function of time, e.g.

s = t^{3} - 6 t^{2} + 9 t

, we use differentiation to find velocity and acceleration, and integration to work backwards.

The Calculus Relationships

The fundamental relationships connect displacement (s), velocity (v), and acceleration (a):

→differentiate

←integrate

Key Formulae

v = \frac{d s}{d t}

a = \frac{d v}{d t} = \frac{d ^{2} s}{d t ^{2}}

v = \int a d t

s = \int v d t

Using Initial Conditions

When we integrate, we get a constant of integration (+c). We find this constant using initial conditions or boundary conditions.

Common Conditions

PhraseMeaning"Starts from rest"When t = 0, v = 0"Starts from the origin O"When t = 0, s = 0"Initially at point where s = 5"When t = 0, s = 5"When t = 2, v = 6"Boundary condition at t = 2

Method

Step 1: Integrate to get expression with +c

Step 2: Substitute the known values (t and s, or t and v)

Step 3: Solve for c

Step 4: Write final expression with c value

Worked Example01

A particle P is moving on the x-axis. At time t seconds, the displacement x metres from O is given by $x = t^{4} - 32 t + 12$ . Find:

(a) the velocity of P when t = 3

(b) the value of t for which P is instantaneously at rest

(c) the acceleration of P when t = 1.5

a)

v = \frac{d x}{d t} = 4 t^{3} - 32

When t = 3:

v = 4 (27) - 32 = 108 - 32 = 76

m/s

b)

At rest means v = 0:

4 t^{3} - 32 = 0

t^{3} = 8

t = 2

seconds

c)

a = \frac{d v}{d t} = 12 t^{2}

When t = 1.5:

a = 12 (1.5)^{2} = 12 \times 2.25 = 27

m/s²

Answers: (a) 76 m/s, (b) t = 2 s, (c) 27 m/s²

Worked Example02

A body moves in a straight line such that its velocity v m/s at time t seconds is given by $v = 2 t^{2} - 16 t + 24$ for $t \geq 0$ . Find:

(a) the initial velocity

(b) the values of t when the body is instantaneously at rest

(c) the greatest speed of the body in the interval $0 \leq t \leq 5$

a)

Initial velocity means t = 0:

v = 2 (0)^{2} - 16 (0) + 24 = 24

m/s

b)

At rest when v = 0:

2 t^{2} - 16 t + 24 = 0

t^{2} - 8 t + 12 = 0

(t - 2) (t - 6) = 0

t = 2

or

t = 6

seconds (only t = 2 in given range)

c)

v is a quadratic with positive t² coefficient, so minimum at turning point.

Turning point at

t = \frac{8}{2} = 4

:

v = 2 (16) - 64 + 24 = - 8

m/s

Check endpoints: v(0) = 24, v(5) = 50 - 80 + 24 = -6

Greatest speed = max(|24|, |-8|, |-6|) = 24 m/s

Answers: (a) 24 m/s, (b) t = 2 s, (c) 24 m/s

Using Integration

When given acceleration or velocity as a function of time, we integrate to work backwards.

Process

Given:

a = f (t)

and initial/boundary conditions

Step 1:

v = \int a d t + c_{1}

Step 2: Use condition to find

c_{1}

Step 3:

s = \int v d t + c_{2}

Step 4: Use condition to find

c_{2}

Common Mistake

Forgetting the constant of integration! Every indefinite integral needs +c. Use the given conditions to find its value.

Worked Example03

A particle is moving on the x-axis. At time t seconds, the velocity of P is $(6 t - t^{2})$ m/s. At t = 0, the particle is at the point where x = 5. Find an expression for the displacement of the particle from O at time t seconds, and the distance of the particle from its starting point when t = 6.

1

Integrate velocity to find displacement:

x = \int (6 t - t^{2}) d t = 3 t^{2} - \frac{t ^{3}}{3} + c

2

When t = 0, x = 5:

5 = 0 - 0 + c

, so

c = 5

Therefore:

x = 3 t^{2} - \frac{t ^{3}}{3} + 5

3

When t = 6:

x = 3 (36) - \frac{216}{3} + 5 = 108 - 72 + 5 = 41

m

4

Distance from starting point = 41 − 5 = 36 m

Answers: $x = 3 t^{2} - \frac{t ^{3}}{3} + 5$ m; Distance = 36 m

Worked Example04

A body starts at rest and moves in a straight line. At time t seconds the displacement of the body from its starting point is s cm, where $s = 4 t^{2} - t^{3}$ for $0 \leq t \leq 4$ .

(a) Show that the body returns to its starting position at t = 4

(b) Find the maximum displacement of the body from its starting point

a)

When t = 4:

s = 4 (16) - 64 = 64 - 64 = 0

✓

The body returns to its starting position.

b)

Maximum displacement when

\frac{d s}{d t} = 0

:

v = \frac{d s}{d t} = 8 t - 3 t^{2} = t (8 - 3 t)

v = 0 when t = 0 or t =

\frac{8}{3}

seconds

3

At

t = \frac{8}{3}

:

s = 4 \times \frac{64}{9} - \frac{512}{27} = \frac{256}{9} - \frac{512}{27} = \frac{768 - 512}{27} = \frac{256}{27}

cm

Answer: Maximum displacement = $\frac{256}{27}$ ≈ 9.48 cm

Variable Acceleration with Vectors

The same calculus rules apply to 2D motion using position vectors. Differentiate and integrate each component separately.

Vector Formulae

If

r = x (t) i + y (t) j

, then:

v = \dot{r} = \overset{x}{˙} i + \overset{y}{˙} j

a = \dot{v} = \overset{x}{¨} i + \overset{y}{¨} j

Key Points

Speed = magnitude of velocity =

∣ v ∣ = \overset{x}{˙}^{2} + \overset{y}{˙}^{2}

When integrating vectors, the constant c is also a vector

Write

c = p i + q j

and use initial conditions to find p and q

Worked Example05

A particle P of mass 0.8 kg is acted on by a single force F N. Relative to a fixed origin O, the position vector of P at time t seconds is r metres, where $r = 2 t^{3} i + 50 t^{- \frac{1}{2}} j$ , $t > 0$ . Find:

(a) the speed of P when t = 4

(b) the acceleration of P as a vector when t = 2

(c) F when t = 2

a)

v = \frac{d r}{d t} = 6 t^{2} i - 25 t^{- \frac{3}{2}} j

When t = 4:

v = 6 (16) i - 25 (4)^{- \frac{3}{2}} j = 96 i - \frac{25}{8} j

Speed =

9 6^{2} + (\frac{25}{8})^{2} = 9216 + 9.77 = 96.1

m/s (3 s.f.)

b)

a = \frac{d v}{d t} = 12 t i + \frac{75}{2} t^{- \frac{5}{2}} j

When t = 2:

a = 24 i + \frac{75}{2} (2)^{- \frac{5}{2}} j = 24 i + 6.63 j

m/s²

c)

F = m a = 0.8 (24 i + 6.63 j) = (19.2 i + 5.30 j)

N (3 s.f.)

Answers: (a) 96.1 m/s, (b) (24i + 6.63j) m/s², (c) (19.2i + 5.30j) N

Worked Example06

A particle P is moving in a plane. At time t seconds, its velocity v m/s is given by $v = 3 t i + \frac{1}{2} t^{2} j$ , $t \geq 0$ . When t = 0, the position vector of P with respect to a fixed origin O is $(2 i - 3 j)$ m. Find the position vector of P at time t seconds.

1

Integrate velocity to find position:

r = \int v d t = \int (3 t i + \frac{1}{2} t^{2} j) d t

r = \frac{3 t ^{2}}{2} i + \frac{t ^{3}}{6} j + c

2

When t = 0,

r = 2 i - 3 j

:

2 i - 3 j = 0 i + 0 j + c

So

c = 2 i - 3 j

3

Therefore:

r = \frac{3 t ^{2}}{2} i + \frac{t ^{3}}{6} j + 2 i - 3 j

r = (\frac{3 t ^{2}}{2} + 2) i + (\frac{t ^{3}}{6} - 3) j

metres

Answer: $r = (\frac{3 t ^{2}}{2} + 2) i + (\frac{t ^{3}}{6} - 3) j$ m

Challenging Example07

A particle P is initially at a fixed origin O. At time t = 0, P is projected from O and moves so that at time t seconds after projection, its position vector r m relative to O is given by $r = (t^{3} - 12 t) i + (4 t^{2} - 6 t) j$ , $t \geq 0$ . Find:

(a) the speed of projection of P

(b) the value of t at the instant when P is moving parallel to j

a)

v = \frac{d r}{d t} = (3 t^{2} - 12) i + (8 t - 6) j

At t = 0 (projection):

v = - 12 i - 6 j

Speed =

(- 12)^{2} + (- 6)^{2} = 144 + 36 = 180 = 65

m/s

b)

Moving parallel to j means the i-component of velocity = 0:

3 t^{2} - 12 = 0

t^{2} = 4

t = 2

seconds (taking positive value)

Answers: (a) $65$ ≈ 13.4 m/s, (b) t = 2 s

Challenging Example08

A particle P moves in a horizontal plane. At time t seconds, the position vector of P is r metres relative to a fixed origin O, and r is given by $r = (20 t - 2 t^{3}) i + k t^{2} j$ , where k is a positive constant. When t = 2, the speed of P is 16 m/s. Find:

(a) the value of k

(b) the acceleration of P at the instant when it is moving parallel to j

a)

v = (20 - 6 t^{2}) i + 2 k t j

When t = 2:

v = (20 - 24) i + 4 k j = - 4 i + 4 k j

Speed = 16, so:

16 + 16 k^{2} = 16

16 + 16 k^{2} = 256

k^{2} = 15

, so

k = 15

(positive)

b)

Moving parallel to j when i-component of v = 0:

20 - 6 t^{2} = 0 \Rightarrow t = \frac{10}{3}

a = \frac{d v}{d t} = - 12 t i + 2 k j = - 12 t i + 215 j

At

t = \frac{10}{3}

:

a = - 12 \frac{10}{3} i + 215 j

a = - 430 i + 215 j

m/s²

Answers: (a) k = √15, (b) a = (−4√30 i + 2√15 j) m/s²

Practice

Question 1: A particle moves in a straight line such that its velocity, v m/s, at time t seconds is given by

v = 3 t^{2} - 10 t + 8

for

t \geq 0

. The particle starts from the origin O. Find the displacement of the particle from O when it first comes to instantaneous rest.

1

At rest when v = 0:

3 t^{2} - 10 t + 8 = 0

(3 t - 4) (t - 2) = 0

t = \frac{4}{3}

or

t = 2

. First rest at

t = \frac{4}{3}

2

s = \int v d t = t^{3} - 5 t^{2} + 8 t + c

At t = 0, s = 0, so c = 0

3

When

t = \frac{4}{3}

:

s = \frac{64}{27} - 5 \times \frac{16}{9} + 8 \times \frac{4}{3} = \frac{64}{27} - \frac{80}{9} + \frac{32}{3}

= \frac{64 - 240 + 288}{27} = \frac{112}{27}

m

Answer: $\frac{112}{27}$ ≈ 4.15 m

Practice

Question 2: At time t seconds, a particle P has position vector

r = (4 t^{2} i + (24 t - 3 t^{2}) j)

metres with respect to a fixed origin O, where

t \geq 0

.

(a) Find the speed of P when t = 2

(b) Show that the acceleration of P is constant and find its magnitude

a)

v = \frac{d r}{d t} = 8 t i + (24 - 6 t) j

When t = 2:

v = 16 i + 12 j

Speed =

1 6^{2} + 1 2^{2} = 256 + 144 = 400 = 20

m/s

b)

a = \frac{d v}{d t} = 8 i - 6 j

This is constant (no t terms).

Magnitude =

8^{2} + (- 6)^{2} = 64 + 36 = 100 = 10

m/s²

Answers: (a) 20 m/s, (b) |a| = 10 m/s²

Practice

Question 3: The velocity of a particle P at time t seconds is

v = ((3 t^{2} - 8) i + 5 j)

m/s. When t = 0, the position vector of P is

(2 i - 4 j)

m. A second particle Q moves with constant velocity

(8 i + 4 j)

m/s. When t = 0, the position vector of Q is

2 i

m. Prove that P and Q collide.

1

For P:

r_{P} = \int (3 t^{2} - 8) i + 5 j d t = (t^{3} - 8 t) i + 5 t j + c

At t = 0,

r_{P} = 2 i - 4 j

, so

c = 2 i - 4 j

r_{P} = (t^{3} - 8 t + 2) i + (5 t - 4) j

2

For Q (constant velocity):

r_{Q} = r_{0} + v t = 2 i + (8 i + 4 j) t

r_{Q} = (8 t + 2) i + 4 t j

3

Collision when

r_{P} = r_{Q}

:

j-components:

5 t - 4 = 4 t \Rightarrow t = 4

Check i-components at t = 4:

P:

64 - 32 + 2 = 34

, Q:

32 + 2 = 34

✓

Both components match at t = 4, so P and Q collide at (34i + 16j) m

Summary

Key Relationships

Operation1D (Scalar)2D (Vector)

Velocity from displacement

v = \frac{d s}{d t}

v = \frac{d r}{d t}

Acceleration from velocity

a = \frac{d v}{d t}

a = \frac{d v}{d t}

Velocity from acceleration

v = \int a d t

v = \int a d t

Displacement from velocity

s = \int v d t

r = \int v d t

Key Exam Skills

Always include +c when integrating (vector c for vectors)

Use initial/boundary conditions to find constants

At rest: v = 0 (or

v = 0

)

Moving parallel to j: i-component of v = 0

Speed = magnitude of velocity:

∣ v ∣ = v_{x}^{2} + v_{y}^{2}

Use F = ma for force problems (F and a are vectors)