The velocity$-$time graph of a car is given below. The car weighs $1000\, kg$.
$(i)$ What is the distance travelled by the car in the first $2$ seconds ?
$(ii)$ What is the braking force at the end of $5$ seconds to bring the car to a stop within one second ?
The distance fravelled by the car in first $2$ seconds $=$ Area of $\Delta ABE$
$=1 / 2 \times AE \times BE$
$=1 / 2 \times 2 \times 15=15 m$
Mass of the car $=1000\, kg$
$F=m \times a,$ where $a=$ slope of the graph $CD$
$a=\frac{0-15}{6-5}=\frac{-15}{1}=-15 m s ^{-2}$
$F =m \times a=1000 \times-15=-15000 N$
How can you find the following ?
$(i)$ Velocity from a displacement$-$time graph.
$(ii)$ Acceleration from velocity$-$time graph.
$(iii)$ Displacement from velocity$-$time graph.
$(iv)$ Velocity from acceleration$-$time graph.
There is an argument about uniform acceleration between $Mr$ $X$ and $Mr$ $Y.$ $Mr$ $X$ says "acceleration means that farther you go faster you go". $Mr$ $Y$ says "acceleration means that longer you go the faster you go". Whose statement is correct ?
$(a)$ When will you say a body is in
$(i)$ uniform motion $(ii)$ non$-$uniform motion ?
$(b)$ Show the path of an object when it is in uniform motion with the help of a graph.
A car starts from rest and moves along the $x-$ axis with constant acceleration $5\, ms^{-2}$ for $8\,\sec $. If it then continues with constant velocity, what distance will the car cover in $12\,\sec $ since it started from the rest ?
A piece of stone is thrown vertically upwards. It reaches its maximum height in $3$ second. If the acceleration of the stone be $9.8\, m s ^{-2}$ directed towards the ground, calculate the initial velocity of the stone with which it is thrown upwards. Find the maximum height attained by it.