## 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 $$(25-20)s$$