A car moving initially at 30m/s comes gradually to a stop in 900m what was the acceleration of the car? I know its -.5 but how?
a = -V²/(2x) = -30²/(2*900) = -0.50 m/s² Units are VERY important
A car moving initially at 30 m/s comes gradually to a stop in 900 m. what was the acceleration of the car?
Given Initial velocity (u) = 30m/sec. Distance (s) = 900m Acceleration (a) = ? Final velocity (v) = 0m/sec Using Kinematic equation:- v² = u² - 2as o² = (30)²-2*a*900 900 = 1800*a 900/1800 = a Acceleration (a) = 900/1800 = 1/2 = 0.5/s² Acceleration (a) = 0.5m/s²
A car moving initially at 30 m/s comes gradually to a stop in 900 m. what was the acceleration of the car?
Given Initial velocity (u) = 30m/sec. Distance (s) = 900m Acceleration (a) = ? Final velocity (v) = 0m/sec Using Kinematic equation:- v² = u² - 2as o² = (30)²-2*a*900 900 = 1800*a 900/1800 = a Acceleration (a) = 900/1800 = 1/2 = 0.5/s² Acceleration (a) = 0.5m/s²
A car accelerates from rest at 2 m/s^2 for 8s, how far has the car traveled in the 7th second?
(a) By s = ut + 1/2at^2 =>s(7) = 0 + 1/2 x 2 x (7)^2 = 49 m =>s(6) = 0 + 1/2 x 2 x (6)^2 = 36 m =>s(7th) = s(7) - s(6) = 49 - 36 = 13 m (b) 1st phase:- =>By s = ut + 1/2at^2 =>s1 = 0 + 1/2 x 2 x (8)^2 = 64 m By v = u + at =>v = 0 + 2 x 8 = 16 m/s 2nd phase:- =>s2 = v x t2 =>s2 = 16 x 20 = 320 m 3rd phase:- By s = ut + 1/2at^2 =>s3 = 16 x 10 + 1/2 x 5 x (10)^2 =>s3 = 410 m By v = u + at =>v = 16 + 5 x 10 =>v = 66 m/s 4th phase:- =>By v^2 = u^2 - 2as =>0 = (66)^2 - 2 x 4 x s4 =>s4 = 544.50 m =>Total distance covered (s) = s1+s2+s3+s4 =>s = 64+320+410+544.50 = 1338.50 m
From the first equation of motion,v=u+at where v=final velocity =0 (car comes to stop)u=15 m/st=10 sWe have, a=(v-u)/t=(0-15)/10=-1.5 m/s^2Which is acceleration (negative value indicates deceleration )From Newton’s second law, FORCE, F=ma where m=mass of carSo, m= F/a=(-3000)/(-1.5) kg=2000 kg
According to the formulav²=u²+2asHere, v=25m/s=>u=20m/s=>a=1.8m/s²=>s=?=>v²=u²+2as=>(25)²=(20)²+2×1.8×s=>625=400+3.6s=>-3.6s=400–625=>-3.6s=-225=>s=-225/-3.6=>s=62.5
Let's put kinematics to play, shall we?Consider this equation: [math]v^2 - u^2 = 2*a*s[/math] ------- (1)Here 'v' is the final velocity of the body in question 'u' is the initial velocity 'a' is the acceleration experienced, and 's' is the distance traveled.Taking only S.I. units, for the case mentioned in the question, it can be calculated that a = -1/8 m/s/s (v = 0, u = 10, s = 400). The negative sign indicates that the body (the car) is decelerating.Applying this result to another equation, [math]v - u = a*t[/math] -------- (2)You can calculate the time taken (t) as t = -10/(-1/8).Which is: t = 80 seconds.PS: 1) The two equations used are two out of the three basic equations of motions in kinematics. These are applicable on of objects seen on a daily basis on macroscopic scales and not accurately on objects at the subatomic or celestial level.2) These equations can simply be derived from the fundamental distance-speed-time relationship. For further details, check: http://www.schoolphysics.co.uk/a...
Using the formula:(final velocity)^2 = (initial velocity)^2 + 2 x (acceleration) x (displacement)(0 m/s)^2 = (5 m/s)^2 + 2 x (acceleration) x (15 m)acceleration = -((5 m/s)^2) / (2 x 15 m) = -0.83 m/s^2From this, we know that a constant deceleration of 0.83 m/s^2 would stop the car in 15 m. Therefore, any deceleration of greater magnitude would stop the car within 15 m.