IB Mathematics

Function	Graph
displacement
acceleration

Question 1

The velocity v in m s⁻¹ of a moving body at time t seconds is given by v = 50 - 10t.

(a) Find the value of its acceleration in m s⁻².

(b) The velocity may also be expressed as `v = (ds)/(dt)` , where s is the displacement in metres.
Given that s = 40 when t = 0, find an expression for s as a function of t.

(a) `a = (dv)/(dt) = -10`

(b) `s = int u dt = 50t - 5t^2 + c`

`40 = 50(0) - 5(0) + c`

`s = 50t - 5t^2 + 40`

Question 2

A car starts by moving from a fixed point A. Its velocity, `v ms^-1` after `t` seconds is given by `v = 4t + 5 - 5e^-t`. Let `d` be the displacement from A when `t = 4` .

(a) Write down an integral which represents `d`.

(b) Calculate the value of `d`.

(a) `d = int_0^4 (4t + 5 - 5e^-t) dt`

(b) `d = [2t^2 + 5t + 5e^-t]_0^4`

`= (32 + 20 + 5e^-4) - (5)`

`= 47 + 5e^-4` (47.1, 3sf)

Question 3

The displacement `s` metres of a car, `t` seconds after leaving a fixed point A , is given by `s = 10t - 0.5t^2`

(a) Calculate the velocity when `t = 0`

(b) Calculate the value of t when the velocity is zero.

(c) Calculate the displacement of the car from A when the velocity is zero.

(a) Velocity is `(ds)/(dt)`

`(ds)/dt = 10 - t`

`10 (m s^-1)`

(b) The velocity is zero when `(ds)/dt = 0`

`10 - t = 0`

`t = 10` (secs)

(c) `s = 50` (metres)

Question 4

The displacement, `s` metres, of a car `t` seconds after it starts from a fixed point A is given by `s = 4t + 5 - 5e^-t`.

(a) Find an expression for its velocity (in `ms^-1` ) after `t` seconds.

(b) Find the acceleration (in `ms^-2` ) at A .

(a) For recognizing the need to differentiate `v = 4 + 5e^-t`

(b) `a = -5e^-t`

At A, `t=0`

`a=-5`

Question 5

The velocity `v` in `ms^-1` of a moving body at time `t` seconds is given by `v=e^(2t-1)`. When `t=0.5` the displacement of the body is 10 m . Find the displacement when `t=1`.

`s = intvdt`

`s = 1/2 e^(2t-1) + c`

Substituting `t=0.5`

`1/2 + c = 10`

`c=9.5`

Substituting `t=1`

`s = 1/2 e + 9.5` (=10.9 to 3s.f.)

Question 6

A particle moves along a straight line so that its velocity, `v``m``s^-1` at time `t` seconds is given by `v=6e^(3t) + 4`. When `t=0`, the displacement, `s` , of the particle is 7 metres. Find an expression for `s` in terms of `t`.

evidence of anti-differentiation

e.g. `s=int(6e^(3x)+4)dx`

`s=2e^(3t)+4t+C`

substituting `t=0`

`7=2+C`

`s=2e^(3t)+4t+5`

Question 7

A particle is moving with a constant velocity along line `L`. Its initial position is A(6, -2, 10). After one second the particle has moved to B(9, -6, 15).

(a) (i) Find the velocity vector, `vec (AB)`.

(ii) Find the speed of the particle.

(b) Write down an equation of the line `L`.

(a) (i) evidence of approach

e.g. `vec(AO) + vec(OB), B - A, ((9-6),(-6+2),(15-10))`

`vec(AB) = ((3),(-4),(5)) " (accept " (3,-4,5) ")"`

(ii) evidence of finding the magnitude of the velocity vector

e.g. speed `= sqrt(3^2 + 4^2 + 5^2)`

(b) correct equation (accept Cartesian and parametric forms)

e.g. `r = ((6),(-2),(10)) + t((3),(-4),(5)) " or " r = ((9),(-6),(15)) + t((3),(-4),(5))`

Question 8

The acceleration, `a` `ms^-2`, of a particle at time `t` seconds is given by `a=2t+cost`.

(a) Find the acceleration of the particle at `t=0`.

(b) Find the velocity, `v`, at time `t`, given that the initial velocity of the particle is `2ms^-1`.

(c) Find `int_0^3vdt`, giving your answer in the form `p-qcos3`.

(d) What information does the answer to part (c) give about the motion of the particle?

(a) substituting `t=0`

e.g. `a(0)=0+cos0`

`a(0)=1`

(b) evidence of integrating the acceleration function

e.g. `int(2t+cost)dt`

correct expression `t^2+sint+c`

evidence of substituting `(0,2)` into indefinite integral

e.g. `2=0+sin0+c,c=2`

`v(t)=t^2+sint+2`

(c) `int(t^2+sint+2)dt=t^3/3-cost+2t`

evidence of using `v(3)-v(0)`

correct substitution

e.g. `(9-cos3+6)-(0-cos0+0), (15-cos3)-(-1)`

`16-cos3` (accept `p=16, q=-1)` )

(d) reference to motion, reference to first 3 seconds

e.g. displacement in 3 seconds, distance travelled in 3 seconds

Question 9

In this question s represents displacement in metres and t represents time in seconds.

The velocity `v``ms^-1` of a moving body is given by `v=40-at` where `a` is a non-zero constant.

(a) (i) If `s=100` when `t=0`, find an expression for `s` in terms of `a` and `t`.

(ii) If `s=0` when `t=0`, write down an expression for `s` in terms of `a` and `t`.

Trains approaching a station start to slow down when they pass a point P. As a train slows down, its velocity is given by `v=40-at`, where `t=0` at P. The station is 500 m from P.

(b) A train M slows down so that it comes to a stop at the station.

(i) Find the time it takes train M to come to a stop, giving your answer in terms of `a`.

(ii) Hence show that `a=8/5`.

(c) For a different train N , the value of `a` is 4 .

Show that this train will stop before it reaches the station.

(a)

(i)

`s = int(40 - at) dt`

`s = 40t - 1/2 at^2 + c`

substituting `s=100` when `t=0`(c=100)

`s = 40t - 1/2 at^2 + 100`

(ii)

`s = 40t - 1/2 at^2`

(b)

(i)

stops at station so `v=0`

`t = 40/a " (seconds)"`

(ii)

evidence of choosing formula for `s` from (a) (ii)

substituting `t=40/a`

e.g. `40xx40/a - 1/2 axx40^2/a^2`

setting up equation

e.g. `500=s, 500=40xx40/a - 1/2 axx40^2/a^2, 500=1600/a-800/a`

evidence of simplification to an expression which obviously leads to `a=8/5`

e.g. `500a = 800, 5 = 8/a, 1000a = 3200-1600`

`a=8/5`

(c)

METHOD 1

`v=40-4t`, stops when `v=0`

`40-4t=0`

`t=10`

substituting into expression for `s`

`s=40xx10-1/2xx4xx10^2`

`s=200`

since `200<500` (allow FT on their `s`, if `s<500` )

train stops before the station

METHOD 2

from (b) `t=40/4=10`