Question

A motorist starts from rest and attains a velocity of 16 m/s2 in 4 sec. Calculate the acceleration and the distance covered by the motorist​

Answer 1

Answer:

acceleration is : 4m/s^2

distance travelled is :32m

Explanation: a=v-u/t

u=0m/s^2

v=16m/s^2

t= 4 sec

a=16-0/4=16/4=4

distance travelled(s):s=ut+1/2at^2

s=0*4+1/2*4*(4)^2

s=0+1/2*4*16

s=4*8=32m

Hope it helps

Answer 2

Provided that:

• Initial velocity = 0 mps
• Final velocity = 16 mps
• Time = 4 seconds

To calculate:

• Acceleration
• Distance

Solution:

• Acceleration = 4 mps sq.
• Distance = 32 m

Using concepts:

✴️ We can use either acceleration formula or first equation of motion to solve this question's part a.

• Choice may vary and yours!

✴️ We can use either third equation of motion or second equation of motion to solve this question's part b.

• Choice may vary and yours!

Using formulas:

✴️ Three equations of motion are mentioned below respectively:

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

✴️ Acceleration formula is mentioned below:

• ${\small{\underline{\boxed{\pmb{\sf{a \: = \dfrac{v-u}{t}}}}}}}$

Where, a denotes acceleration, u denotes initial velocity, v denotes final velocity, t denotes time taken and s denotes displacement or distance or height.

Choice may vary and yours!

Required solution:

✡️ Firstly let us find out the acceleration by using first equation of motion!

${\sf{:\implies v \: = u \: + at}}$

${\sf{:\implies 16 = 0 + a(4)}}$

${\sf{:\implies 16 = 4a}}$

${\sf{:\implies a \: = 4 \: ms^{-2}}}$

${\sf{:\implies Acceleration \: = 4 \: ms^{-2}}}$

✡️ Now let us find out the acceleration by using acceleration formula!

$:\implies \sf a \: = \dfrac{Change \: in \: velocity}{Time} \\ \\ :\implies \sf a \: = \dfrac{dv}{dt} \\ \\ :\implies \sf a \: = \dfrac{\Delta v}{t} \\ \\ :\implies \sf a \: = \dfrac{v - u}{t} \\ \\ :\implies \sf a \: = \dfrac{16 - 0}{4} \\ \\ :\implies \sf a \: = \dfrac{16}{4} \\ \\ :\implies \sf a \: = 4 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = 4 \: ms^{-2}$

• Henceforth, acceleration or rate in change in velocity is 4 mps sq.

(Dear user, I have calculated the acceleration by using two months, you can use any one of them!)

Choice may vary and yours!

✡️ Now by using third equation of motion let us calculate the distance!

$:\implies \sf v^2 - u^2 \: = 2as \\ \\ :\implies \sf (16)^{2} - (0)^{2} = 2(4)(s) \\ \\ :\implies \sf 256 - 0 = 8s \\ \\ :\implies \sf 256 = 8s \\ \\ :\implies \sf \dfrac{256}{8} \: = s \\ \\ :\implies \sf 32 \: m \: = s \\ \\ :\implies \sf s \: = 32 \: m \\ \\ :\implies \sf Distance \: = 32 \: m$

✡️ Now by using second equation of motion let us find out the distance travelled!

$:\implies \sf s \: = ut + \: \dfrac{1}{2} \: at^2 \\ \\ :\implies \sf s \: = 0(4) + \dfrac{1}{2} \times 4(4)^{2} \\ \\ :\implies \sf s \: = 0 + \dfrac{1}{2} \times 4(16) \\ \\ :\implies \sf s \: = 0 + \dfrac{1}{2} \times 64 \\ \\ :\implies \sf s \: = 32 \: m \\ \\ :\implies \sf Distance \: = 32 \: m$

• Henceforth, distance travelled = 32 metres!

(Dear user, I have calculated the distance by using two months, you can use any one of them!)

Choice may vary and yours!

Hope it's helpful! :)