A motor bike running at $90 \,km h ^{-1}$, is slowed down to $54 \,km h^{-1}$ by the application of brakes, over a distance of $40\, m$. If the brakes are applied with the same force, calculate $(i)$ total time in which bike comes to rest $(ii)$ total distance travelled by bike.
Case $1$ : $u=90 km h ^{-1}=(-) 25 ms ^{-1} ; v=54 km$
$h^{-1}=15 m \cdot s ^{-1} ; S =40 m ; a=?$
Applying $v^{2}-u^{2}=2 a S$
$(15)^{2}-(25)^{2}=2 \times a \times 40$
$80 a=-400$
$a=-5 m s ^{-2}$
Case $2$ : $u=90 km h ^{-1}=25 m s ^{-1} ; v=0 ; S =?$
$a=-5 m s ^{-2} ; t=?$
Applying $v^{2}-u^{2}=2 a S$
$(0)^{2}-(25)^{2}=2 \times(-5) \times S$
$S=\frac{-625}{-10}=62.5 m$
Applying $\quad v=u+a t$
$0=25-5 \times t$
or $5 t=25$ or $t=5 s$
A body thrown in the vertically upward direction rises upto a height $'h^{\prime}$ and comes back to the position of its start.
Calculate :
$(a)$ the total distance travelled by the body and
$(b)$ the displacement of the body. Under what condition will the magnitude of the displacement be equal to the distance travelled by an object ?
Two cars moving in opposite directions cover same distance $'d'$ in one hour. If the cars were moving in north$-$south direction, what will be their displacement in one hour ?
Velocity$-$time graph for the motion of an object in a straight path is a straight line parallel to the time axis.
$(a)$ Identify the nature of motion of the body.
$(b)$ Find the acceleration of the body.
$(c)$ Draw the shape of distance-time graph for this type of motion.
What is the relationship between the distance travelled and the time elapsed for motion with uniform velocity ?
$(a)$ Derive second equation of motion $S=u t+\frac{1}{2} a t^{2}$ graphically where the symbols have their usual meanings.
$(b)$ A car accelerates uniformly from $18\, km h ^{-1}$ to $36\, km h^{-1}$ in $5$ seconds. Calculate the acceleration and the distance covered by the car in that time.