Question

A truck starts from rest and accelerates uniformly at 3 m/s2 . At t=10 sec a stone is dropped by a person standing at top of truck ((7 m high from ground) What are velocity and acceleration of stone at t=11 sec.(g=10 m/s2 , neglect air resistance)
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Answer 1

Given:

• Initial velocity of truck , u = 0 [ start from rest ]
• Acceleration of the truck , a = 3 m/s²
• At = 10 sec , a stone is dropped by a person standing at top of truck , [ 7 m high from the ground ]

To Find :

• Velocity of the stone
• and acceleration of the stone

Solution :

We have ,

Intital velocity of truck , u = 0

Acceleration ,a = 3m/ s²

Time , t = 10 sec

By Equation of motion

$\sf\:v=u+at$

$\sf\implies\:v=0+3\times10$

$\sf\implies\:v=30ms^{-1}$

Thus , Final velocity of the truck is 30 m / s

At t = 10 sec , a stone is dropped by a person standing at top of truck.

Thus , Velocity of the stone at t = 10 sec is 30m/s

We have to Find the velocity and acceleration of the stone at = 11 sec .

1) We have to find velocity of stone

The Horizontal component of the velocity of stone , remains unchanged because no air resistance and any other force in x - axis .

$\sf\:v_{x}=30ms^{-1}$

The vertical component of the velocity of stone :

We have ,

$\sf\:u_y=0ms^{-1}$

$\sf\:a_y=10ms^{-1}$

$\sf\:t=11-10=1sec$

By Equation of motion:

$\sf\:v_y=u_y+a_y\times\:t$

$\sf\implies\:v_y=0+10$

$\sf\implies\:v_y=10ms^{-1}$

Net Velocity of stone :

$\sf\:v=\sqrt{(v_x)^2+(v_y)^2}$

$\sf\implies\:v=\sqrt{(30)^2+(10)^2}$

$\sf\implies\:v=\sqrt{900+100}$

$\sf\implies\:v=\sqrt{1000}$

$\sf\implies\:v=31.6ms^{-1}$

Hence , Velocity of stone at t=11 s is 31.6 m/s

2) We have to find acceleration of stone :

When the stone is dropped , there is no force in x - axis , stones moves only the under influence of gravity .

Hence , the acceleration of stone is 10 m/s² in vertically downward direction .

_______________

${\purple{\boxed{\large{\bold{Formula's}}}}}$

Kinematic equations for uniformly accelerated motion.

$\bf\:v=u+at$

$\bf\:S=ut+\frac{1}{2}at{}^{2}$

$\bf\:v{}^{2}=u{}^{2}+2aS$

and $\bf\:s_{nth}=u+\frac{a}{2}(2n-1)$

Answer 2

$\Large\bf{\color{indigo}GiVeN\:}$

• A truck starts from rest.

• Accelerates uniformly at $\bf{3\:m/s^2\:}$.

• Time (t) = 10 s

$\bf\red{We\:know\:that,}$

$\orange\bigstar\:\:\bf{\color{lime}v_f\:=\:v_i\:+\:at\:}$

$\bf\pink{Where,}$

• $\bf{v_f\:=\:Final\:velocity}$

• $\bf{v_i\:=\:0\:m/s^2}$

$\longmapsto\:\:\bf{v_f\:=\:0\:+\:3\times{10}\:}$

$\longmapsto\:\:\bf{v_f\:=\:30\:m/s}$

$\bf{\color{coral}Therefore,}$

The final velocity of the truck is 30m/s.

And The velocity of the stone is 30m/s.

☆ At 't = 11s' the horizontal component $\bf{(v_x)}$ of velocity, in the absence of air resistance, remains unchanged, i.e.

$\longmapsto\:\:\bf\purple{v_x\:=\:30\:m/s\:}$

☆ The vertical component $\bf{(v_y)}$ of velocity of the stone is given by the first equation of motion as;

$\green\bigstar\:\:\bf\blue{v_y\:=\:u\:+\:a_y{t}\:}$

$\bf\pink{Where,}$

• $\bf{a_y\:=\:g\:=\:10\:m/s^2}$

• $\bf{v_i\:=\:0\:m/s^2}$

• t = 11 - 10 = 1 s

$\longmapsto\:\:\bf{v_y\:=\:0\:+\:10\times{1}\:}$

$\longmapsto\:\:\bf\purple{v_y\:=\:10\:m/s}$

☆ The resultant velocity (v) of the stone is given as;

$\red\checkmark\:\:\bf\pink{v\:=\:\sqrt{v_x^2\:+\:v_y^2}\:}$

$\longmapsto\:\:\bf{v\:=\:\sqrt{30^2\:+\:10^2}\:}$

$\longmapsto\:\:\bf{v\:=\:\sqrt{900\:+\:100}\:}$

$\longmapsto\:\:\bf{v\:=\:\sqrt{1000}\:}$

$\longmapsto\:\:\bf{v\:=\:10\sqrt{10}\:m/s}$

$\longmapsto\:\:\bf\green{v\:=\:31.62\:m/s}$

✒ Let be the angle made by the resultant velocity with the horizontal component of velocity, $\bf{v_x}$ be $\bf{\theta\:}$.

$:\implies\:\:\bf{\color{olive}tan{\theta}\:=\:\dfrac{v_y}{v_x}\:}$

$:\implies\:\:\bf{\theta\:=\:\tan^{-1}\Big (\dfrac{v_y}{v_x}\Big)\:}$

$:\implies\:\:\bf{\theta\:=\:\tan^{-1}\Big (\dfrac{10}{30}\Big)\:}$

$:\implies\:\:\bf{\theta\:=\:\tan^{-1}\Big (\dfrac{1}{3}\Big)\:}$

$:\implies\:\:\bf\blue{\theta\:=\:18.26\:}$

♕ When the stone is dropped from the truck, the horizontal force acting on it becomes zero. However, the stone continues to move under the influence of gravity.

♕ Hence, the acceleration of the stone is $\bf\orange{10\:m/s^2}$ and it acts vertically downward.