This document will allow university Math students to practice velocity, displacement and acceleration questions using differentiation and integration. All the questions are similar in style and full worked solutions have been provided. This will allow you to understand how a certain question in thi...
Velocity, displacement and acceleration questions involving
differentiation and integration
1) The particle M starts from rest at the point O on a line. The particle M travels
in a straight line. The acceleration of M is 𝑎 = 9𝑡 − 45 at time 𝑡 𝑠 once it has
left O.
Calculate the distance M travels before it comes to instantaneous rest.
𝑑𝑣
𝑎= 𝑑𝑡
𝑎 = 9𝑡 − 45
𝑑𝑣
𝑑𝑡
= 9𝑡 − 45
𝑣 = ∫9𝑡 − 45 𝑑𝑡
9 2
𝑣= 2
𝑡 − 45𝑡 + 𝐶1
Calculating value of 𝐶1
9 2
Sub 𝑣 = 0 and 𝑡 = 0 in 𝑣 = 2
𝑡 − 45𝑡 + 𝐶1
9 2
0= 2
(0) − 45(0) + 𝐶1
𝐶1 = 0
9 2 9 2
Substitute 𝐶1 = 0 back into 𝑣 = 2
𝑡 − 45𝑡 + 𝐶1 and we get 𝑣 = 2
𝑡 − 45𝑡
𝑑𝑠
𝑣= 𝑑𝑡
(𝑠 is the displacement)
𝑑𝑠 9 2
𝑑𝑡
= 2
𝑡 − 45𝑡
𝑑𝑠
Substitute 𝑑𝑡
= 0 to find maximum and minimum values
9 2
2
𝑡 − 45𝑡 = 0
9
𝑡( 2 𝑡 − 45) = 0
𝑡=0
9
2
𝑡 − 45 = 0
𝑡 = 10
𝑡 = 10 is the time when M comes to rest after having travelled
We will now calculate the displacement when 𝑡 = 10
𝑑𝑠 9 2
𝑑𝑡
= 2
𝑡 − 45𝑡
9 2
𝑠= ∫ 2 𝑡 − 45𝑡 𝑑𝑡
3 3 45 2
𝑠= 2
𝑡 − 2 𝑡 + 𝐶2
Calculating value of 𝐶2
, 3 3 45 2
Sub 𝑠 = 0 and 𝑡 = 0 in 𝑠 = 2
𝑡 − 2
𝑡 + 𝐶2
3 3 45 2
0= 2
(0) − 2
(0) + 𝐶2
𝐶2 = 0
3 3 45 2 3 3 45 2
Substitute 𝐶2 = 0 back into 𝑠 = 2
𝑡 − 2
𝑡 + 𝐶2 and we get 𝑠 = 2
𝑡 − 2
𝑡
3 3 45 2
Substitute 𝑡 = 10 in 𝑠 = 2
𝑡 − 2
𝑡
3 3 45 2
𝑠= 2
(10) − [ 2
(10) ]
𝑠 =− 750 (Displacement can be a negative number)
Distance travelled by M before it comes to instantaneous rest = 750m
FINAL ANSWER
750m
, 2) The particle R starts from rest at the point O on a line. The particle R travels in
a straight line. The acceleration of R is 𝑎 = 7𝑡 − 23 at time 𝑡 𝑠 once it has left
O.
Calculate the distance R travels before it comes to instantaneous rest.
𝑑𝑣
𝑎= 𝑑𝑡
𝑎 = 7𝑡 − 23
𝑑𝑣
𝑑𝑡
= 7𝑡 − 23
𝑣 = ∫7𝑡 − 23 𝑑𝑡
7 2
𝑣= 2
𝑡 − 23𝑡 + 𝐶1
Calculating value of 𝐶1
7 2
Sub 𝑣 = 0 and 𝑡 = 0 in 𝑣 = 2
𝑡 − 23𝑡 + 𝐶1
7 2
0= 2
(0) − 23(0) + 𝐶1
𝐶1 = 0
7 2 7 2
Substitute 𝐶1 = 0 back into 𝑣 = 2
𝑡 − 23𝑡 + 𝐶1 and we get 𝑣 = 2
𝑡 − 23𝑡
𝑑𝑠
𝑣= 𝑑𝑡
(𝑠 is the displacement)
𝑑𝑠 7 2
𝑑𝑡
= 2
𝑡 − 23𝑡
𝑑𝑠
Substitute 𝑑𝑡
= 0 to find maximum and minimum values
7 2
2
𝑡 − 23𝑡 = 0
7
𝑡( 2 𝑡 − 23) = 0
𝑡=0
7
2
𝑡 − 23 = 0
46
𝑡= 7
46
𝑡= 7
is the time when R comes to rest after having travelled
46
We will now calculate the displacement when 𝑡 = 7
𝑑𝑠 7 2
𝑑𝑡
= 2
𝑡 − 23𝑡
7 2
𝑠= ∫ 2 𝑡 − 23𝑡 𝑑𝑡
7 3 23 2
𝑠= 6
𝑡 − 2 𝑡 + 𝐶2
Calculating value of 𝐶2
7 3 23 2
Sub 𝑠 = 0 and 𝑡 = 0 in 𝑠 = 6
𝑡 − 2
𝑡 + 𝐶2
, 7 3 23 2
0= 6
(0) − 2
(0) + 𝐶2
𝐶2 = 0
7 3 23 2 7 3 23 2
Substitute 𝐶2 = 0 back into 𝑠 = 6
𝑡 − 2
𝑡 + 𝐶2 and we get 𝑠 = 6
𝑡 − 2
𝑡
46 7 3 23 2
Substitute 𝑡 = 7
in 𝑠 = 6
𝑡 − 2
𝑡
7 46 3 23 46 2
𝑠= 6
( 7
) − [ 2
( 7
)]
𝑠 =− 165. 537415 (Displacement can be a negative number)
Distance travelled by R before it comes to instantaneous rest = 165.54m (2dp)
FINAL ANSWER
165.54m (2dp)
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