Question

A car travelling
with a
velocity of 10 m
is stopped by applying brakes to produce a uniform retardation of 1.25 ms^-2 calculate
the distance by the car before
it come
to rest. Also calculate the time taken by
car to stop​

Answer 1

Answer:

PLEASE MAKE IT BRAINLIEST.

v^2 - u^2 = 2as

v=0 ( as car stops) u = 10m/s a= -1.25 ( retardation is negative)

thetefore

u^2 = 2as

100 = 2 ×1.25 × s

s= 100/2.5

= 40m

Time taken

v = u + at

v=0 u=10 a= -1.25

t= 10/1.25

= 8 sec

Answer 2

Gɪᴠᴇɴ

Initial velocity of car = 10 m/s

Retardation of car = 1.25 m/s²

Tᴏ ꜰɪɴᴅ

Distance travelled by car before coming to rest & time taken to stop.

Sᴏʟᴜᴛɪᴏɴ ❶ 

As the car comes to rest, so final velocity of car (v) = 0 m/s

Again, retardation of car means negative acceleration.

So, acceleration of car = -1.25 m/s²

Now using equation of motion :

⇒ v² = u² + 2as

⇒ 0² = 10² + 2(-1.25)s

⇒ 0 = 100 - 2.5s

⇒ 100 = 2.5s

⇒ s = 100/2.5

⇒ s = 40 m

Hᴇɴᴄᴇ,

Distance travelled by car is 40 m

Sᴏʟᴜᴛɪᴏɴ ❷

Here, we know equation of motion :

⇒ a = (v - u)/t

⇒ -1.25 = (0 - 10)/t

⇒ -1.25t = -10

⇒ t = -10/-1.25

⇒ t = 8 seconds.

Hᴇɴᴄᴇ,

Time taken by car to stop is 8 seconds.