Question

A car is travelling with a velocity of 40 m/s.
Calculate the time taken by the car to
come to rest if it is given a uniform
retardation of 5 m/s by apPlying breaks.
Calculate displacement of the car
during this time as well.
​

Answer 1

Explanation:

Initial velocity of car-40m/s

Final velocity of car-0m/s

Retardation-5m/s square

Time-?

By 1st equation of motion

v=u+at

0=40+5×t

Transposing 40 to LHS

-40=5×t

Transposing -40 to RHS and 5t to LHS

5t=40(it became positive because of transposing)

t=40/5=8 sec

Sorry I can't help with displacement

Answer 2

Answer:

$\bigstar{\bold{Time\:taken\:to\:come\:to\:rest=8\:s}}$

$\bigstar{\bold{Displacement=160\:m}}$

Explanation:

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

• Initial velocity (u) = 40 m/s
• Final velocity (v) = 0 m/s
• Acceleration (a) = 5 m/s²

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

• Time taken (t)
• Displacement (s)

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

→ Here we have to find the time taken by the car to come to rest.

→ By the first equation of motion,

  v = u + at

→ Substituting the given data,

  0 = 40 + -5 × t

→ Here acceleration is taken as negative since it is retardation or deceration.

 -40 = -5t

  t = 40/5

  t = 8 s

→ Hence time taken by the car is 8 s

  $\boxed{\bold{Time\:taken\:to\:come\:to\:rest=8\:s}}$

→ Now we have to find the dispacement or distance travelled by the car

→ By the third equation of motion,

  v² - u² = 2as

→ Substitute the data,

  0² - 40² = 2 × -5 × s

  -1600 = -10s

   s = 1600/10

   s = 160 m

→ Hence the displacement of the car is 160 m

$\boxed{\bold{Displacement=160\:m}}$

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

→ The three equation of motion are:

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