Question

Brakes applied to a
car produce negative acceleration
by 10m/s. If
car takes
5 second's to stop after
"applying brakes , calculate distance covered by car before
coming to rest
.​

Answer 1

$\large\underline{\underline{\red{\sf \purple{\maltese}\:\: Given :- }}}$

$\sf\implies Brakes \ produce \ a \ negative \ accl^n \ of \ 10m/s^2 .\\\sf\implies The \ car \ took \ 5 s \ to\ stop .$

$\large\underline{\underline{\red{\sf \purple{\maltese}\:\: To \ Find :- }}}$

$\sf\implies The \ distance \ travelled \ by \ the \ car .$

$\large\underline{\underline{\red{\sf \purple{\maltese}\:\: Solution :- }}}$

According to the question the brakes applied byba car produce a negative acceleration of 10m/s² . The car comes to rest in 5s . Since the car comes to rest it's final velocity will be 0m/s .

$\underline{\purple{\boldsymbol{ Using \ 1st \ equation \ of \ motion .}}}$

$\sf:\implies \green{ v = u + at }\\\\\sf:\implies 0m/s = u + (-10m/s^2) 5 \\\\\sf:\implies u = 0 + 50m/s \\\\\sf:\implies\boxed{\pink{\sf u = 50m/s }}$

________________________________

$\underline{\purple{\boldsymbol{ Using \ 2nd \ equation \ of \ motion .}}}$

$\sf\green{:\implies s = ut + \dfrac{1}{2}at^2}\\\\\sf:\implies s =( 50m/s)(5s) + \dfrac{1}{2}\times (-10m/s^2)\times (5s)^2 \\\\\sf:\implies s = 250m - 5(25) m \\\\\sf:\implies s = 250m - 125 m \\\\\sf:\implies\boxed{\pink{\sf s = 125m }}$

$\boxed{\blue{\bf Hence \ the \ distance \ travelled \ is 125 m.}}$

Answer 2

Distance = 375m.

Explanation:

Given -

• Retardation(negative acceleration) = - 10m/s²
• Final velocity (v) = 0m/s
• Time (t) = 5 seconds

Solution -

Initial velocity -

• v= u+at
• 0= u+ (-10×5)
• 0 = u-50
• 50= u

Initial velocity is 50m/s.

Distance -

• ut+1/2at²
• (50×5) +1/2×(10)(5)²
• 250+ 250/2
• 250+ 125
• 375

Distance is 375m.