Study the speed-time graph of a car given below and answer the questions that follow:
$(i)$ What type of motion is represented by $OA$?
(ii) What type of motion is represented by $AB$?
(iii) What type of motion is represented by $BC$?
(iv) What is the acceleration of the car from $O$ to $A$?
$(v)$ What is the acceleration of the car from $A$ to $B$?
(vi) What is the retardation of the car from $B$ to $C$?
$(i)$ What type of motion is represented by $OA$?
(ii) What type of motion is represented by $AB$?
(iii) What type of motion is represented by $BC$?
(iv) What is the acceleration of the car from $O$ to $A$?
$(v)$ What is the acceleration of the car from $A$ to $B$?
(vi) What is the retardation of the car from $B$ to $C$?
(N/A) $(i)$ The speed-time graph is a straight line inclined to the time axis from $O$ to $A$. This represents a uniformly accelerated motion.
(ii) The speed-time graph is a straight line parallel to the time axis from $A$ to $B$. This represents uniform motion (constant speed).
(iii) The speed-time graph is a straight line from $B$ to $C$ having a negative slope. This represents a uniformly retarded motion.
(iv) Change in speed $= 40 - 0 = 40 \ m \ s^{-1}$.
Change in time $= 10 - 0 = 10 \ s$.
Acceleration $a = \frac{\text{Change in speed}}{\text{Change in time}} = \frac{40}{10} = 4 \ m \ s^{-2}$.
$(v)$ Since the motion is uniform (constant speed),the acceleration $a = 0 \ m \ s^{-2}$.
(vi) Change in speed $= 0 - 40 = -40 \ m \ s^{-1}$.
Change in time $= 50 - 30 = 20 \ s$.
Retardation $= -(\text{Acceleration}) = -(\frac{0 - 40}{50 - 30}) = -(\frac{-40}{20}) = -(-2) = 2 \ m \ s^{-2}$.
(ii) The speed-time graph is a straight line parallel to the time axis from $A$ to $B$. This represents uniform motion (constant speed).
(iii) The speed-time graph is a straight line from $B$ to $C$ having a negative slope. This represents a uniformly retarded motion.
(iv) Change in speed $= 40 - 0 = 40 \ m \ s^{-1}$.
Change in time $= 10 - 0 = 10 \ s$.
Acceleration $a = \frac{\text{Change in speed}}{\text{Change in time}} = \frac{40}{10} = 4 \ m \ s^{-2}$.
$(v)$ Since the motion is uniform (constant speed),the acceleration $a = 0 \ m \ s^{-2}$.
(vi) Change in speed $= 0 - 40 = -40 \ m \ s^{-1}$.
Change in time $= 50 - 30 = 20 \ s$.
Retardation $= -(\text{Acceleration}) = -(\frac{0 - 40}{50 - 30}) = -(\frac{-40}{20}) = -(-2) = 2 \ m \ s^{-2}$.
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