Question

A cheetah can accelerate from rest at the rate of 4m/s a) What will be the velocity attained by it in 10s b) How far will it travel in this duration​

Answer 1

Answer:

$\bigstar{\bold{Final\:velocity=40\:m/s}}$

$\bigstar{\bold{Distance\:covered=200\:m}}$

Explanation:

$\Large{\underline{\sf{Given:}}}$

• Acceleration = 4 m/s²
• Initial velocity = 0 m/s
• Time taken = 10 s

$\Large{\underline{\rm{To\:Find:}}}$

• Velocity attained by it in 10 s
• Distance travelled during this duration

$\Large{\underline{\sf{Solution:}}}$

Final velocity:

↬ First we have to find the final velocity of the cheetah.

↬ By the first equation of motion we know that,

   v = u + at

↬ Substituting the values we get,

    v = 0 + 4 × 10

   v = 40 m/s

↬ Hence the velocity attained by the cheetah is 40 m/s.

    $\boxed{\bold{Final\:velocity=40\:m/s}}$

Distance travelled:

↬ Now we have to find the distance travelled during this duration.

↬ By the third equation of motion,

   v² - u² = 2as

↬ Substitute the data,

    40² - 0² = 2 × 4 × s

    8s = 1600

      s = 1600/8

      s = 200 m

↬ Hence the distance covered by the cheetah is 200 m.

   $\boxed{\bold{Distance\:covered=200\:m}}$

$\Large{\underline{\sf{Notes:}}}$

↬ The three equations of motion are:

• v = u + at
• s = ut + 1/2 × a × t²
• v² - u² = 2as

where v = final velocity

           u = initial velocity

           a = acceleration

           t = time taken

           s = distance covered

Answer 2

Explanation:

Given:-

initial velocity =u=0m/s

acceleration=a=4m/s

time taken =t=10s

To find :-

Final velocity =v

distance covered =s

Solution:-

as we know that

• According to first equation of motion

${:}\longrightarrow$${\boxed{v=u+at}}$

[By putting the values]

${:}\longrightarrow$$v=0+4×10$

${:}\longrightarrow$${\underline {\boxed{\bf {v=40m/s}}}}$

Now,

• According to second equation of motion

${:}\longrightarrow$${\boxed{s=ut+{\frac {1}{2}}at {}^{2}}}$

[by putting the values]

${:}\longrightarrow$$s=0×10+{\frac {1}{{\cancel{2}}}}×{\cancel{4}}×10$

${:}\longrightarrow$$s=10×20$

${:}\longrightarrow$${\underline{\boxed{\bf {s=200m}}}}$

$\therefore$${\boxed{Final\:velocity=40m /s}}$

${\boxed {Distance\:covered=200m}}$