The following graph describes the motion of a girl going to meet her friend who stays $50\, m$ from her house
$(i)$ How much time she takes to reach her friend's house ?
$(ii)$ What is the distance travelled by the girl during the time$-$interval $0$ to $12$ minute ?
$(iii)$ During which time-interval she is moving towards her house ?
$(iv)$ For how many minutes she was at rest, during the entire journey ?
$(v)$ Calculate the speed by which she returned home.
$(i)$ $14$ minute.
$(ii)$ $40 m +20 m =60 m$.
$(iii)$ Between $14$ to $16$ minutes.
$(iv)$ $2+2+2=6$ minutes.
$(v)$ Speed $=\frac{\text { Total distance covered }}{\text { Total time taken }}$
$=\frac{50 m }{2 min }=25 m \min ^{-1}$
$=\frac{25 \times 60}{1000} kmh ^{-1}=1.5 km h ^{-1}$
The graph given below is the distance$-$time graph of an object.
$(i)$ Find the speed of the object during first four seconds of its journey.
$(ii)$ How long was it stationary ?
$(iii)$ Does it represent a real situation ? Justify your answer.
Mention the nature of motion of a body if its displacement$-$time graph is a straight line parallel to time axis.
The brakes applied to a car produce an acceleration of $6\, m s ^{-2}$ in the opposite direction of motion. If the car takes $2\, s$ to stop after the application of the brakes, calculate the distance it travels during this time.
Draw a diagram to show the motion of a body whose speed remains constant but velocity continuously changes.
Under what conditions of motion, the distance covered by a body is equal to the magnitude of its displacement ?