The distance$-$time graph of two trains are given below. The trains start simultaneously in the same direction.
$(i)$ How much ahead of $A$ is $B$ when the motion starts ?
$(ii)$ What is the speed of $B$ ?
$(iii)$ When and where $A$ will catch $B$ ?
$(iv)$ What is the difference between the speeds of $A$ and $B$ ?
$(v)$ Is the speed of either trains uniform or non uniform ? Justify your answer.
$(i)$ $100\, km$
$(ii)$ Speed of $B=\frac{O R}{P R}=\frac{150-100}{2-0}=\frac{50}{2}$
$=25 km h ^{-1}$
$(iii)$ They at a distance of $150\, km$ after $2$ hours.
$(iv)$ Speed of $A =\frac{150-0}{2-0}=\frac{150}{2}=75 km h ^{-1}$
Speed of $B=25 km h ^{-1}$
So the difference is $75-25=50 km h ^{-1}$
$(v)$ Uniform as the distance-time graph is a straight line.
A train $100 \,m$ long is moving with a velocity of $60\, km h^{-1}$. Find the time it takes to cross the bridge $1\, km$ long.
The position$-$time graph for the motion of a car is given below
$(i)$ How far the car tavelled in the time interval $2$ to $6 s ?$
$(ii)$ During which interval of time its speed was more?
$(iii)$ Calculate the average speed of the car.
A hiker rides $700\, m$ north, $300$ meast, $400 \,m$ north $600\, m$ west, $1200\, m$ south, $300\, m$ east and finally $100\, m$ north. Draw the path of motion of the biker. What distance did he cover ? What was his displacement ?
Using following data, draw time-displacement graph for a moving object :
Time $(s)$ | $0$ | $2$ | $4$ | $6$ | $8$ | $10$ | $12$ | $14$ | $16$ |
Displacement $(m)$ | $0$ | $2$ | $4$ | $4$ | $4$ | $6$ | $4$ | $2$ | $0$ |
Use this graph to find average velocity for first $4\,\sec $, for next $4\,\sec $ and for last $6\,\sec $.
A satellite while revolving around the earth completes one revolution in $1$ hour and $30$ minutes. What is the angular speed of the satellite ?