- Home
- Standard 9
- Science
The $v-t$ graph of cars $A$ and $B$ which start from the same place and move along straight road in the same direction, is shown. Calculate
$(i)$ the acceleration of car $A$ between $0$ and $8\, s$.
$(ii)$ the acceleration of car $B$ between $2\, s$ and $4\, s$.
$(iii)$ the points of time at which both the cars have the same velocity.
$(iv)$ which of the two cars is ahead after $8\, s$ and by how much ?
Solution
$(i)$ Acceleration of car $A$ between $0$ to $8\, s$ is
$a=\frac{\text { speed }}{\text { time }}=\frac{80}{8}=10 m s ^{-2}$
$(ii)$ Acceleration of car $B$ between $2$ to $4\, s$ is
$a=\frac{\text { speed }}{\text { time }}=\frac{(60-20)}{2}=20 m s ^{-2}$
$(iii)$ After $2\, s$ and $6\, s$ from the start.
$(iv)$ Distance travelled by $CMA$
Area of $\Delta OAB =\frac{1}{2} \times 80 \times 8$
$=320 m$
Distance travelled by car $B$
Area of the trapezium $CDEB$
$=\frac{1}{2} \times(7+4) \times 60$
$=330 m$
Car $B$ is ahead by $10\, m$.
