Question

A car of mass 1000 kg is moving with velocity 20m/s suddenly apply brakes and come to rest in 5 seconds. Calculate distance travelled by car before coming to rest.

Answer 1

m=1000kg
u=20m/s
v=0m/s
t=5s
a=(v-u)/t
a=0-20/5=-4m/s^2
F=ma=-4000N
2as=v^2-u^2
s=(0m/s)^2-(20m/s)^2/-8
s=(-400m^2/s^2)÷(-48m/s^2)
s=50m

Answer 2

$\red{HEY\:BUDDY!!}$

HERE'S THE ANSWER..

______________________________

♠️ From the given question we will calculate following data , it has applied break so it's final Velocity will be 0

⏺️ Final velocity ( v ) = 0

⏺️ Initial velocity ( u ) = 20 m / s

⏺️ Time ( t ) = 5 second

⏺️ Acceleration ( a ) = ?

✔️ Using first Eqn of motion $\bold{v\:=u\:+a\:×\:t}$

=> v = u + a × t

=> a = ( v - u ) / t

=> a = ( 0 - 20 ) / 5

=> a = - 20 / 5

=> $\boxed{ a\:=\:-4 \: m/s^2}$

✔️ Here Acceleration is negative be car is stopping or retarding

⏺️ Now using third Eqn of motion

✔️ $\bold{v^2\:=u^2\:+2\:×a\:×\:s}$

⏺️ Where ( s ) is displacement

=> ( v )^2 = ( u )^2 + 2 × a × s

=> 0 = ( u )^2 + 2 × a × s

=> ( 20 )^2 + 2 ( - 4 ) ( s ) = 0

=> - 8 ( s ) = - 400

=> s = 400 / 8

=> $\boxed{ s\:=\:50 \: m}$

♠️ After applying the brake car will travel 50 meter .

HOPE HELPED..

$\red{JAI HIND..}$

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