Question

A tourist drops a rock from rest from a guard rail overlooking a valley. What is the velocity of the rock at 4.0 s? What is the displacement of the rock at 4.0 s

Answer 1

GiveN :

• Initial velocity of the rock, u = 0 m/s

• Time taken, t = 4 sec

• Acceleration, a = 10 m/s²

⠀⠀⠀⠀⠀⠀⠀⠀⠀

We've to find out the velocity of rock after 4 second's & the displacement of rock after 4 seconds.

⠀⠀⠀⠀⠀⠀⠀⠀⠀

We'll find the velocity of rock after 4 seconds by 1st equation of motion & the displacement of rock after 4 seconds by 2nd equation of motion.

⠀⠀⠀⠀⠀⠀⠀⠀⠀

We know that,

$\bigstar \: \large{ \boxed{ \frak{v = u + at}}}$

where we've the values as :

• v = unknown.
• u = 0 m/s
• a = 10 m/s²
• t = 4 sec

⠀⠀⠀⠀⠀⠀⠀⠀⠀

Substituting the values :

⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\longrightarrow\sf v = 0 + 10 \times 4$

$\longrightarrow\sf v = 0 + 40$

$\longrightarrow \large{ \boxed{\sf{ v = 40m/s}}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\therefore{\underline{\sf{ Velocity \: of \: rock \: after \: 4 \: seconds \: is\textsf{ \textbf{40 m/s}}}}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀

Now, By using Second equation of motion :

⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\bigstar \: { \boxed{ \frak{ s = ut + \dfrac{1}{2}at^2}}}$

where we've the values,

• v = 40 m/s
• u = 0 m/s
• a = 10 m/s²
• t = 4 sec

⠀⠀⠀⠀⠀⠀⠀⠀⠀

Substituting the values :

⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\longrightarrow\sf s = 0 \times 10 + \dfrac{1}{2} \times 10 \times 4^2$

$\longrightarrow\sf s = 0 + 5 \times 16$

$\longrightarrow\large{ \boxed{ \sf{s = 80 m. }}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\therefore{ \underline{ \sf{Displacement \: of \: the \: rock \: after \: 4 \: seconds \: is \: \textsf{ \textbf{80 m/s}}}}}$