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 $.
Average velocity for first $4\,\sec $
Average velocity $=\frac{\text { change in displacement }}{\text { Total time taken }}$
$v=\frac{4-0}{4-0}=\frac{4}{4}=1 \,ms ^{-1}$
For next $4\sec $ , $v=\frac{4-4}{8-4}=\frac{0}{4}=0 \,ms ^{-1}$
(or as $x$ remains the same from $4$ to $8\,\sec $, velocity is zero)
For last $6\,\sec , v=\frac{0-6}{16-10}=-\,1\,m s^{-1}$
Define 'average speed'. An object moves with a uniform speed of $10\, m s ^{-1}$ for $5 s$ and then with a uniform speed of $5\, m s ^{-1}$ for $10\, s$. Find its average speed.
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.
What is the nature of motion of a particle depicted by following displacement$-$time graphs ?
Can a particle be accelerated
$(i)$ if its speed is constant ?
$(ii)$ if its velocity is constant ?
A person travelling in a bus noted the timings and the corresponding distances as indicated on the km stones. (a) Name this type of table $(b)$ What conclusion do you draw from this data ?
Time | Distance |
$8.00\, am$ | $10\, km$ |
$8.15 \,am$ | $20 \,km$ |
$8.30\, am$ | $30\, km$ |
$8.45\, am$ | $40\, km$ |
$9.00\, am$ | $50\, km$ |