Velocity, displacement and acceleration questions with integration and differentiation

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)