Question

a car is moving with a velocity of 72 km/hr . On the application of brakes , car stops in 10 seconds .
Find 1. retardation.....
2. distance covered by the car before it stops .....
3. braking force if mass of car is 800 kg.....




need answer urgent..
plz help​

Answer 1

Answer:

u= 72km/hr=72x5/18=20km/hr

t= 10s

v=0

v=u+at

0=20+10a

-20=10a

a = -2m/s^2

s=ut + 1/2 at^2

s= 20x10+1/2x-2x100

s= 200-100

s=100m

f=ma

f=800x-2

f=-1600 kg m/s

Hope it helps

Answer 2

✪AnSwEr

${\tt{\red{\underline{\large{Given}}}}}$

• A. car moving with
• velocity :-72km/h
• car stop in 10 s
• After break

${\sf{\green{\underline{\large{To\:find}}}}}$

• retardation.....
• 2. distance covered by the car before it stops .....
• 3. braking force if mass of car is 800 kg

${\sf{\pink{\underline{\Large{Explanation}}}}}$

72km/h

• changing it into ms

by multiplying by 5/18

NOW

=72 *5/18

u=20m/s

From equation of motion

v=u+at

=>v=20+10a

=>0=20+10a

=>-2=a

From equation of motion

v²=u²+2as

=>0=400-4s

=>-400=-4s

=>s=100

Now force

force = mass × acceleration

=>f=800×-2

=>f=-1600