A body starts from rest and is uniformly accelerated for 30s, The distance traveled in the first 10s is x,next 10s is y and the last 10s is z Then x:y:z is the same as
Answers
Answer:
Answer:
\fbox {a = - 1.5 m/s^2}
a=−1.5m/s
2
\fbox {Distance (s) = 300m}
Distance(s)=300m
===========================
Given data :-
Initial velocity (u) = 30m/s
Final velocity (v) = 0
Since the body comes to rest after applying breaks , so final velocity is " zero "
Time taken :- 20 s
=============================
Question :- To find deceleration and distance covered
===============================
FOR Solving Problem Like THESE We Use 3 Equations Of Uniform Acceleration .They Are
(1) V = u + at
(2) v^2 - u^2 = 2as
(3) S = ut + 1/2 at^2
We use equation (1) to find deceleration because the rest 2 equations has " s " in it ( s is unknown value )
\fbox {V = u + at}
V=u+at
0 = 30 + a (20)
30 + 20a =0
20 a = -30
a = -30/20
a = - 3/2
a = -1.5 m/s^2
Now it's time to calculate " distance"
\fbox {V^2 - u^2 = 2as}
V
2
−u
2
=2as
0 - (30)^2 = 2 (-1.5)s
- 900 = -3S
900 = 3s
S = 900/3
S = 300m
◇ THING TO KNOW :-
Acceleration in negative direction is called as deceleration.
Answer:
1:3:5
Explanation:
*let the acceleration of the body is a and u = 0
*then x = 1/2at^2=1/2a(10)^2
*y=1/2a(20)^2-x
*y=1/2a(20)^2-1/2a(10)^2
*y=1/2a(10)(30)
*z=1/2a(30)^2-1/2a(20)^2
*z=1/2a(10)(50)
*******x:y:z=1:3:5*******
I hope this may help you