Question

a car is moving with a velocity of 45m/s comes to rest in 10 second. Find the retardation produced.Also calculate the distance covered by the car during this period​

Answer 1

Explanation:

initial value=45m/s

final value=0m/s

time=10sec

retardation=v-u/t

=0-45/10

=45/10

=4.5m/s

S=ut+1/2at2

=45*10+1/2* 4.5*10*10

=450+225

=675 (Ans)

Answer 2

Answer:

a) retardation=4.5m/s^2

b)the distance traveled during this time=225m

Explanation:

We know that the car stops after 10 secs.

Therefore

u=45m/s   /initial velocity

v=0            /final velocity

t=10secs

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a)w.k.t

 v=u+(a*t)-----/eq. of motion

 substituting values

 0=45+(a*10)

 -45=a*10

 -45/10=a

 a= -4.5m/s^2

-----------------------------------------------------------

b)now we have a= -4.5m/s^2

 w.k.t $v^{2}-u^{2}=2as$

substituting values

(0)^2-(45)^2=2*(-4.5)*s

-2025= -2*4.5*s

-2025 =s

   -9

s=225m

thats all..