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Solve problems involving speed-time graphs
Speed-Time Graphs
7.2
Speed-Time Graphs
In a speed-time graph
the vertical axis represents the speed of a motion
the horizontal axis represents the time taken
the gradient of graph represents the
rate of change of speed with respect to time,
that is
acceleration
A
speed-time graph
can be drawn if following information regarding the motion is obtained
Speed-time table
Equation that represents the relationship between speed and time
Example
\(P\)
The speed of object increases
from
\(0 \space ms^-1\)
to
\(16 \space ms^-1\)
The gradient of graph is positive, hence the rate of change of speed is positive
\(\text{Acceleration} = \dfrac{\text{Change of speed}}{\text{Changein time}}\)
The area of triangle
\(OAP\)
that is the area represents the
distance
travelled in the period of
\(10s\)
\(Q\)
There is
no change of speed
(zero gradient)
The object moves at a
uniform speed
The area of square under
\(Q\)
represents the
distance
travelled in the period of
\((20-10)s\)
\(R\)
Speed of the object decreases
The gradient of graph is negative, hence the rate of change of speed is negative
\(\text{Deceleration} = \dfrac{\text{Change of speed}}{\text{Change in time}}\)
There is no change in direction in which the motion of the object remains in the same direction
The area of triangle
\(BCR\)
that is the area represents the distance travelled in the period of
\((25-20)s\)
Speed-Time Graphs
7.2
Speed-Time Graphs
In a speed-time graph
the vertical axis represents the speed of a motion
the horizontal axis represents the time taken
the gradient of graph represents the
rate of change of speed with respect to time,
that is
acceleration
A
speed-time graph
can be drawn if following information regarding the motion is obtained
Speed-time table
Equation that represents the relationship between speed and time
Example
\(P\)
The speed of object increases
from
\(0 \space ms^-1\)
to
\(16 \space ms^-1\)
The gradient of graph is positive, hence the rate of change of speed is positive
\(\text{Acceleration} = \dfrac{\text{Change of speed}}{\text{Changein time}}\)
The area of triangle
\(OAP\)
that is the area represents the
distance
travelled in the period of
\(10s\)
\(Q\)
There is
no change of speed
(zero gradient)
The object moves at a
uniform speed
The area of square under
\(Q\)
represents the
distance
travelled in the period of
\((20-10)s\)
\(R\)
Speed of the object decreases
The gradient of graph is negative, hence the rate of change of speed is negative
\(\text{Deceleration} = \dfrac{\text{Change of speed}}{\text{Change in time}}\)
There is no change in direction in which the motion of the object remains in the same direction
The area of triangle
\(BCR\)
that is the area represents the distance travelled in the period of
\((25-20)s\)
Chapter : Motion Graph
Topic : Solve problems involving speed-time graphs
Form 4
Mathematics
View all notes for Mathematics Form 4
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