MATH 31
UNIT 4 LESSON #2
ACCELERATION
NAME ANSWERS
Page 1 of 5
Average Acceleration (slope of secant line on velocity graph)
𝒂𝒂𝒗𝒆 =
𝒗(𝒕𝟐 ) − 𝒗(𝒕𝟏 )
𝒕𝟐 − 𝒕𝟏
Instantaneous acceleration is the first derivative of the velocity function v(t) or the
second derivative of the position function s(t)
𝑣(𝑡) = 𝑠
′ (𝑡)
𝑑𝑠
=
𝑑𝑡
𝑑𝑣
𝑎(𝑡) = 𝑣′(𝑡) =
𝑑𝑡
𝑎(𝑡) = 𝑠 ′′ (𝑡) =
𝑑2𝑠
𝑑𝑡 2
EXAMPLE 1: The velocity (in m/s) of an object is given by v (t )= 3t2 – 5 t + 8,
where t > 0. Find the acceleration at t = 3.
v(t ) = 3t2 – 5t + 8
𝒅𝒗
a(t) = 𝒅𝒕 = 6t – 5
a(3) = 6(3) – 5 = 13 m/s2
EXAMPLE 2: A particle is moving in a straight line. The distance from a fixed
point is given by s(t ) = t3 – 5 t2 + 7t + 9, where t > 0.
a) Determine the velocity and acceleration at any time.
v(t) = s = 3t2 – 10t + 7
𝒅𝒗
a(t ) = 𝒅𝒕 = 6t – 10
b) Determine the turning points: that is, the points where the velocity is
zero.
v (t )= 3t2 – 10t + 7 = 0
(3t – 7 )(t – 1 ) = 0
t = 7/3 or t = 1
s(7/3) = (7/3) 3 – 5 (7/3) 2 + 7(7/3) + 9= 10 22/27
s(1) = (1) 3 – 5 (1) 2 + 7(1) + 9= 12
Therefore, (1, 12) and (7/3, 10 22/27) are the turning points.
U4 L2 ANS Acceleration
MATH 31
UNIT 4 LESSON #2
ACCELERATION
NAME ANSWERS
Page 2 of 5
EXAMPLE 2 continued
c) Draw the graph of s(t) = t3 – 5 t2 + 7t + 9
d) Draw the position diagram.
t=0
s=9
t = 2 1/3
s =1022/27
t=1
s = 12
c) Determine the total distance covered in the first 5 seconds.
Distance during 1st second is 12 – 9 = 3 m
Distance from 1 s to 2 1/3 s is 12 – 10 22/27 = 1 5/27
Distance from 2 1/3 s to 5 s is 44 – 10 22/27 = 33 5/27
Total distance = 3 + 1 5/27 + 33 5/27 = 37 10/27
U4 L2 ANS Acceleration
t=5
s = 44
MATH 31
UNIT 4 LESSON #2
ACCELERATION
NAME ANSWERS
Page 3 of 5
QUESTIONS
1. In each of these questions, the velocity of a particle is given. Find the acceleration
at any time and at the particular times required.
a) v(t) = 2t – 3 t2
b) v(t) = (t + 3)2
/8
a(t) = 2 – 6t
a(t) = 2(t + 3)(1) = 2t + 6
a(4) = 2 – 6(4) = - 22
a(3) = 2(3) + 6 = 12
c) v(t) = 5t3 – 2t
a(t) = 15t2 – 2
a(2) = 15(2)2 – 2 = 58
d)
𝑣(𝑡) =
10𝑡
3 + 𝑡2
(𝟑 + 𝒕𝟐 )(𝟏𝟎) − 𝟏𝟎𝒕(𝟐𝒕)
𝒂(𝒕) =
(𝟑 + 𝒕𝟐 )𝟐
𝟐
−𝟏𝟎𝒕 + 𝟑𝟎)
𝒂(𝒕) =
(𝟑 + 𝒕𝟐 )𝟐
−𝟏𝟎(𝟏)𝟐 + 𝟑𝟎) 𝟓
𝒂(𝟏) =
=
(𝟑 + (𝟏))𝟐
𝟒
2. An object is moving in a straight line. the displacement (in metres) from a fixed
point is given by s(t) = t2 – 5 t + 4, where t > 0.
a) Find the velocity and acceleration at any time.
v(t) = 2t – 5
a(t) = 2
/4
b) Determine the turning point; that is, the point where the velocity is zero.
2t – 5 = 0
t = 2.5
s = (2.5)2 – 5 (2.5)+ 4 = - 2.25
The turning point is (2.5, 2.25)
U4 L2 ANS Acceleration
MATH 31
UNIT 4 LESSON #2
ACCELERATION
NAME ANSWERS
Page 4 of 5
3. The position of a braking car at any time t is s(t) = 28t – 0.2t3. The position is
measured in metres and the time in seconds.
a) Determine the acceleration when t = 1s.
/5
𝒗(𝒕) =
𝒅𝒔
= 𝟐𝟖 − 𝟎. 𝟔𝒕𝟐
𝒅𝒕
𝒂(𝒕) =
𝒅𝟐 𝒔 𝒅𝒗
=
= −𝟏. 𝟐𝒕
𝒅𝒕𝟐 𝒅𝒕
𝒂(𝟏) = −𝟏. 𝟐(𝟏) = −𝟏. 𝟐 𝒎/𝒔𝟐
b) How long does it take the car to stop?
𝟐𝟖 − 𝟎. 𝟔𝒕𝟐 = 𝟎
𝟐𝟖 = 𝟎. 𝟔𝒕𝟐
𝒕𝟐 = 𝟒𝟔.
𝒕 = 𝟔. 𝟖 𝒔
4. Determine the velocity of a particle when its acceleration is zero if its position in
metres at time t seconds is given by s(t) = 2t3 – 48t2.
𝒅𝒔
𝒗(𝒕) =
= 𝟔𝒕𝟐 − 𝟗𝟔𝒕
𝒅𝒕
/4
𝒅𝟐 𝒔 𝒅𝒗
𝒂(𝒕) = 𝟐 =
= 𝟏𝟐𝒕 − 𝟗𝟔
𝒅𝒕
𝒅𝒕
𝟏𝟐𝒕 − 𝟗𝟔 = 𝟎
𝒕=𝟖𝒔
𝒗(𝟖) = 𝟔(𝟖)𝟐 − 𝟗𝟔(𝟖) = −𝟑𝟖𝟒 𝒎/𝒔
5. One snowmobile was following another machine that stalled 15 m directly ahead
of it. The brakes are applied immediately and the snowmobiler’s position is
determined by s(t) = 12t – t3, t > 0, where t is in seconds and s(t) in metres. Will a
crash occur?
𝒗(𝒕) =
𝒅𝒔
= 𝟏𝟐 − 𝟑𝒕𝟐
𝒅𝒕
𝟐
/4
𝟏𝟐 − 𝟑𝒕 = 𝟎
𝟑𝒕𝟐 = 𝟏𝟐
𝒕𝟐 = 𝟒
𝒕 = ±𝟐
U4 L2 ANS Acceleration
s(2) = 12(2) – (2)3 = 16 metres
Yes there will be a crash!!!
(Why do we not consider the root t = – 2 ?)
MATH 31
UNIT 4 LESSON #2
ACCELERATION
NAME ANSWERS
U4 L2 ANS Acceleration
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