Question

0-6. A ball starts from rest and rolls down 16m down an inclined plane in 4 s. 2mark ta, Wing is the acceleration of the ball? b) What is the velocity of the ball at the bottom of the incline?​

Answer 1

Provided that:

• Initial velocity = 0 mps
• Distance = 16 metres
• Time = 4 seconds

Don't be confused! Initial velocity cames as zero because the ball starts from rest.

To calculate:

• Acceleration
• Final velocity

Solution:

• Acceleration = 2 mps sq.
• Final velocity = 8 mps

Using concepts:

• If we are finding acceleration first then we have to use second equation of motion.

• We can use either third equation of motion or first equation of motion to find out the final velocity.

Using formulas:

• The below mentioned are three equations of motion respectively.

$\begin{gathered}\boxed{\begin{array}{c}\\ {\pmb{\sf{Three \: equations \: of \: motion}}} \\ \\ \sf \star \: v \: = u \: + at \\ \\ \sf \star \: s \: = ut + \: \dfrac{1}{2} \: at^2 \\ \\ \sf \star \: v^2 - u^2 \: = 2as\end{array}}\end{gathered}$

Required solution:

~ Firstly let us find out the acceleration by using second equation of motion!

$:\implies \sf s \: = ut + \: \dfrac{1}{2} \: at^2 \\ \\ :\implies \sf 16 = 0(4) + \dfrac{1}{2} \times a \times (4)^{2} \\ \\ :\implies \sf 16 = 0 + \dfrac{1}{2} \times a \times 16 \\ \\ :\implies \sf 16 = 0 + \dfrac{1}{2} \times 16a \\ \\ :\implies \sf 16 = \dfrac{1}{2} \times 16a \\ \\ :\implies \sf 16 = \dfrac{1}{\cancel{{2}}} \times \cancel{16}a \\ \\ :\implies \sf 16 = 1 \times 8a \\ \\ :\implies \sf 16 = 8a \\ \\ :\implies \sf \dfrac{16}{8} \: = a \\ \\ :\implies \sf \cancel{\dfrac{16}{8}} \: = a \\ \\ :\implies \sf 2 \: = a \\ \\ :\implies \sf a \: = 2 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = 2 \: ms^{-2}$

~ Now let us find out the final velocity by using either third equation of motion or first equation of motion!

By using first equation of motion:

$:\implies \sf v \: = u \: + at \\ \\ :\implies \sf v \: = 0 + 2(4) \\ \\ :\implies \sf v \: = 0 + 8 \\ \\ :\implies \sf v \: = 8 \: ms^{-1} \\ \\ :\implies \sf Final \: velocity \: = 8 \: ms^{-1}$

By using third equation of motion:

$:\implies \sf v^2 - u^2 \: = 2as \\ \\ :\implies \sf v^2 - (0)^{2} = 2(2)(16) \\ \\ :\implies \sf v^2 = 2(2)(16) \\ \\ :\implies \sf v^2 = 4(16) \\ \\ :\implies \sf v^2 = 64 \\ \\ :\implies \sf v \: = \sqrt{64} \\ \\ :\implies \sf v \: = 8 \: ms^{-1} \\ \\ :\implies \sf Final \: velocity \: = 8 \: ms^{-1}$

• Choice may yours to use either first equation of motion or third equation of motion to find the final velocity.