Question 1:
A car moves from rest with an acceleration of 2.0 m s–2. Calculate:
(a) velocity of car after 5.0 s.
(b) distance travelled in 5.0 s.
(c) distance travelled in the fifth second.
Answer:
$$ \begin{aligned} u & =0 \mathrm{~ms}^{-1} \\ a & =2.0 \mathrm{~ms}^{-2} \\ t & =5.0 \mathrm{~s} \\ v & =? \end{aligned} $$
(a)
$$ \begin{aligned} v & =u+a t \\ & =0+(2.0)(5.0) \\ & =10.0 \mathrm{~m} \mathrm{~s}^{-1} \end{aligned} $$
(b)
$$ \begin{aligned} s & =u t+\frac{1}{2} a t^2 \\ s_1 & =0+\frac{1}{2}(2.0)(5.0)^2 \\ & =25.0 \mathrm{~m} \end{aligned} $$
(c)
$$ \text { Fifth second means the time between } 4 \mathrm{~s} \text { and } 5 \mathrm{~s} \text {. } $$
$$ \begin{aligned} & \text { For } t=4 \mathrm{~s} \\ & s_2=0+\frac{1}{2}(2.0)(4.0)^2 \\ & =16.0 \mathrm{~m} \end{aligned} $$
$$ \begin{aligned} &\text { Therefore, distance travelled in the fifth second }\\ &\begin{aligned} s & =s_1-s_2 \\ & =25.0-16.0 \\ & =9.0 \mathrm{~m} \end{aligned} \end{aligned} $$
A car moves from rest with an acceleration of 2.0 m s–2. Calculate:
(a) velocity of car after 5.0 s.
(b) distance travelled in 5.0 s.
(c) distance travelled in the fifth second.
Answer:
$$ \begin{aligned} u & =0 \mathrm{~ms}^{-1} \\ a & =2.0 \mathrm{~ms}^{-2} \\ t & =5.0 \mathrm{~s} \\ v & =? \end{aligned} $$
(a)
$$ \begin{aligned} v & =u+a t \\ & =0+(2.0)(5.0) \\ & =10.0 \mathrm{~m} \mathrm{~s}^{-1} \end{aligned} $$
(b)
$$ \begin{aligned} s & =u t+\frac{1}{2} a t^2 \\ s_1 & =0+\frac{1}{2}(2.0)(5.0)^2 \\ & =25.0 \mathrm{~m} \end{aligned} $$
(c)
$$ \text { Fifth second means the time between } 4 \mathrm{~s} \text { and } 5 \mathrm{~s} \text {. } $$
$$ \begin{aligned} & \text { For } t=4 \mathrm{~s} \\ & s_2=0+\frac{1}{2}(2.0)(4.0)^2 \\ & =16.0 \mathrm{~m} \end{aligned} $$
$$ \begin{aligned} &\text { Therefore, distance travelled in the fifth second }\\ &\begin{aligned} s & =s_1-s_2 \\ & =25.0-16.0 \\ & =9.0 \mathrm{~m} \end{aligned} \end{aligned} $$
Question 2:
Encik Nizam drives a car at a speed of 108 km h–1. Suddenly he sees a car in front moving very slowly. Th erefore, Encik Nizam slows down his car to a speed of 72 km h–1. The displacement made by the car is 125 m. If the acceleration of the car is uniform, calculate
(a) acceleration of Encik Nizam’s car
(b) time taken for the speed of the car to reduce from 108 km h–1 to 72 km h–1.
Answer:
$$ \begin{aligned} &\text { Change speed in unit } \mathrm{km} \mathrm{~h}^{-1} \text { to } \mathrm{m} \mathrm{~s}^{-1} \text {. }\\ &\begin{aligned} u & =108 \mathrm{~km} \mathrm{~h}^{-1} \\ & =\frac{108 \times 1000}{3600} \\ & =30 \mathrm{~m} \mathrm{~s}^{-1} \end{aligned}\\ &\begin{aligned} v & =72 \mathrm{~km} \mathrm{~h}^{-1} \\ & =\frac{72 \times 1000}{3600} \\ & =20 \mathrm{~m} \mathrm{~s}^{-1} \end{aligned} \end{aligned} $$
(a)
$$ \begin{aligned} v^2 & =u^2+2 a s \\ 2 a s & =v^2-u^2 \\ a & =\frac{v^2-u^2}{2 \mathrm{~s}} \\ a & =\frac{20^2-30^2}{2 \times 125} \\ & =-2.0 \mathrm{~m} \mathrm{~s}^{-1} \end{aligned} $$
$$ \text { The car experiences an acceleration of }-2.0 \mathrm{~m} \mathrm{~s}^{-2} \text {. } $$
(b)
$$ \begin{aligned} & u=30 \mathrm{~m} \mathrm{~s}^{-1} \\ & v=20 \mathrm{~m} \mathrm{~s}^{-1} \\ & a=-2.0 \mathrm{~m} \mathrm{~s}^{-1} \end{aligned} $$
$$ \begin{aligned} v & =u+a t \\ t & =\frac{v-u}{a} \\ t & =\frac{20-30}{-2.0} \\ & =5 \mathrm{~s} \end{aligned} $$
Encik Nizam drives a car at a speed of 108 km h–1. Suddenly he sees a car in front moving very slowly. Th erefore, Encik Nizam slows down his car to a speed of 72 km h–1. The displacement made by the car is 125 m. If the acceleration of the car is uniform, calculate
(a) acceleration of Encik Nizam’s car
(b) time taken for the speed of the car to reduce from 108 km h–1 to 72 km h–1.
Answer:
$$ \begin{aligned} &\text { Change speed in unit } \mathrm{km} \mathrm{~h}^{-1} \text { to } \mathrm{m} \mathrm{~s}^{-1} \text {. }\\ &\begin{aligned} u & =108 \mathrm{~km} \mathrm{~h}^{-1} \\ & =\frac{108 \times 1000}{3600} \\ & =30 \mathrm{~m} \mathrm{~s}^{-1} \end{aligned}\\ &\begin{aligned} v & =72 \mathrm{~km} \mathrm{~h}^{-1} \\ & =\frac{72 \times 1000}{3600} \\ & =20 \mathrm{~m} \mathrm{~s}^{-1} \end{aligned} \end{aligned} $$
(a)
$$ \begin{aligned} v^2 & =u^2+2 a s \\ 2 a s & =v^2-u^2 \\ a & =\frac{v^2-u^2}{2 \mathrm{~s}} \\ a & =\frac{20^2-30^2}{2 \times 125} \\ & =-2.0 \mathrm{~m} \mathrm{~s}^{-1} \end{aligned} $$
$$ \text { The car experiences an acceleration of }-2.0 \mathrm{~m} \mathrm{~s}^{-2} \text {. } $$
(b)
$$ \begin{aligned} & u=30 \mathrm{~m} \mathrm{~s}^{-1} \\ & v=20 \mathrm{~m} \mathrm{~s}^{-1} \\ & a=-2.0 \mathrm{~m} \mathrm{~s}^{-1} \end{aligned} $$
$$ \begin{aligned} v & =u+a t \\ t & =\frac{v-u}{a} \\ t & =\frac{20-30}{-2.0} \\ & =5 \mathrm{~s} \end{aligned} $$
Question 3:
Swee Lan rows a boat forward. She uses an oar to push the water backwards. Why is the boat able to move forward?
Answer:
When Swee Lan paddles a boat backwards, a force of action F is applied on the river water and simultaneously a force of reaction of the same magnitude but in the opposite direction to the direction of F’ acts on the boat. Therefore, the boat moves forward.
Swee Lan rows a boat forward. She uses an oar to push the water backwards. Why is the boat able to move forward?
Answer:
When Swee Lan paddles a boat backwards, a force of action F is applied on the river water and simultaneously a force of reaction of the same magnitude but in the opposite direction to the direction of F’ acts on the boat. Therefore, the boat moves forward.