Explanation
Let after a time $$t$$, the cyclist overtake the bus. Then
$$96+\displaystyle\dfrac { 1 }{ 2 } \times 2\times { t }\displaystyle^{ 2 }=20\times t$$ or $${ t }\displaystyle^{ 2 }-20t+96=0$$
$$\therefore \quad t=\displaystyle\dfrac { 20\pm \sqrt { 400-4\times 96 } }{ 2\times 1 }$$
$$\quad \quad =\displaystyle\dfrac { 20\pm 4 }{ 2 } =8 sec$$ and $$12 sec$$
But since they won't meet again after they have crossed each other once, so only $$8$$ seconds is the answer since they are headed in different directions.
Let $$a$$ be the constant acceleration of the particle. Then$$s=ut+\displaystyle\dfrac { 1 }{ 2 } a{ t }\displaystyle^{ 2 }$$ or $${ s }_{ 1 }=0+\displaystyle\dfrac { 1 }{ 2 } \times a\times { \left( 10 \right) }\displaystyle^{ 2 }=50a$$and $${ s }_{ 2 }=\left[ 0+\displaystyle\dfrac { 1 }{ 2 } a{ \left( 20 \right) }^{ 2 } \right] -50a=150a$$
$$\therefore \quad { s }_{ 2 }=3{ s }_{ 1 }$$
Alternatively:Let $$a$$ be constant acceleration and
$$s=ut+\displaystyle\dfrac { 1 }{ 2 } a{ t }\displaystyle^{ 2 }$$, then $${ s }_{ 1 }=0+\displaystyle\dfrac { 1 }{ 2 } \times a\times 100=50a$$
Velocity after 10 sec. is $$v=0+10a$$
So, $${ s }_{ 2 }=10a\times 10+\displaystyle\dfrac { 1 }{ 2 } a\times 100=150a\Rightarrow { s }_{ 2 }=3{ s }_{ 1 }$$
Let $$a$$ be constant acceleration, using $$s=ut+\displaystyle\dfrac { 1 }{ 2 } a{ t }\displaystyle^{ 2 }$$
so distance coreved in first 10 seconds $${ s }_{ 1 }=0+\displaystyle\dfrac { 1 }{ 2 } \times a\times 100=50a$$
So, distance covered in next 10 seconds $${ s }_{ 2 }=10a\times 10+\displaystyle\dfrac { 1 }{ 2 } a\times 100=150a\Rightarrow { s }_{ 2 }=3{ s }_{ 1 }$$
The ball is thrown up and it returns in $$6 sec$$,
Time of accent = Time of decent
So, time to reach the highest point $$=6/2 = 3 s$$
By 2nd equation of motion
$$s = ut + \dfrac{1}{2}gt^{2}$$
To calculate height, consider motion of the ball from highest point to the ground
$$H = \dfrac{1}{2}gt^{2} = \dfrac{1}{2} \times 10 \times 3^{2} =45\ m$$
To calculate projection velocity, consider motion of the ball from projection to return to the point of projection.
$$0=ut-\dfrac{1}{2}gt^2$$
$$u=10 \times 6/2=30\ m/s$$
$$200=u\times 2+\dfrac{1}{2}a(2)^{2}$$
$$100=u+a..........(i)$$
$$420=u(6)+\dfrac{1}{2}a(6)^{2}$$
$$70=u+3a..........(ii)$$
from $$(i)$$ and $$(ii)$$ $$2a=-30\Rightarrow a=-15cm/s^{2}$$
$$\therefore u=100-a=100-(-15)=115\;cm/s$$
$$u_{7^{th}}=u+at=115+(-15)(7)=115-105=10\;cm/sec$$
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