Physics, asked by arohiraj01p7kre6, 1 year ago

a body is moved with a speed of 36 km per hour is brought to rest in 10 seconds what is the negative acceleration and the distance travelled by the body before coming to rest

Answers

Answered by mohitgurung626
1
u(initial velocity)=10km/s
v(final velocity)=0km/s(since distance travelled by the body come to rest
as you know the acceleration is rate of change of velocity.
v=u+at
 \frac{v - u}{t}  = a
\frac{0 - 10}{10}  = a
(-1)=am/s²

and s
 s =  10 \times 10+  \frac{1}{2}  ( - 1) \times 100

a =50m
I hope it works.


Answered by MrBest
1
Hii Friend

Good evening

Lets solve This problem

Q.=Easy

Solution:

Given

Initial velocity=36km/h=(36×15/8)=67.5m/s

Final velocity(v)=0 (Brakes applied and body comes in rest)

Time taken=10sec

Use 1st equation of motion

V=u+at

=>0=67.5+a×10

=>-6.75=a


To find Distance(s) use 3rd equation of motion

2as=v2-u2

2×(-6.75)s=0-(-67.5)sq

2×(-6.75)s=-4556.25

s=-4556.25/2×(-6.75)=337.5m

Answer=Negative acceleration=(-6.75)m/s2

Distance =337.5m


Answered by Mr.BEST

See YOU soon

TILL then stay happy stay best.

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