If a car goes 50 meters in 12 seconds, what is the acceleration of the car?

Answer 1

s=1/2at^2
50=1/2a*144
a=2*50/144 m/s/s
if it accelerates fronm rest uniformly
a car can also go 50 meters at uniform velocity of 50/12 m/s in which case the acceleration is 0
the acceleration may not be uniform
overall the question is an example of data insufficiency

Answer 2

Acceleration should be zero. Since the start and ending speed of the car was not stated, it traveled 50 metres in 12 seconds which is about 164ft in 12 seconds = 13.66ft/sec = 9.3mph

Answer 3

a million there are 1000 milligrams in a kilogram so 500 kg = 500 x 1000 2. 20 km/hr are what proportion metres consistent with 2d? 20 km is 20 x 1000 meters in 60 x 60 seconds so the respond is 20 x 1000 / (60 x 60) 3 30 meters in 2.35 seconds so in a million 2d it is going 30/2.35 meteres consistent with 2d (purely iver 6) 4 you're able to try this now....

Answer 4

getfkdup is wrong acceleration is the change in velocity over time.
you are confusing speed with acceleration.

anyway assumption have to be made as your qns is not clear enough like the initial vel. (the car could be travelling at const vel). therefore i assum that the car starts from rest.

Using s = ut + 0.5at^2
s = 50, u = 0, t = 12
we get, 50 = 72a
a = 72/50
a = 0.694ms^ -2

Answer 5

s=1/2 a t^2 this implies a= 2s /t^2=2*50/12*12=100/144=.694444
units is metres per seconds squared

Answer 6

Vi = initial velocity
Vf = final velocity
Vavg = average velocity
T = time
A = acceleration
D = distance

i am assuming the car starts at complete rest
Vavg = D/T
50 meters / 12 seconds = 4.167m/s
Vavg = (Vf - Vi)/2
or Vf = 2Vavg + Vi
4.167m/s X 2 - 0m/s = 8.333m/s
Vf = 8.333m/s
A = (Vf - Vi)/T
(8.333m/s +0m/s)/12s
A = .0694m/s^2

Answer 7

use this formula. s= ut+1/2at^2 (I think you should know this formula)

u=initial velocity=0(beginning is stop)
a=acceleration
t=time=12 seconds
s=displacement=50 metre

In the end, the answer is 0.6944ms^-2.

Answer 8

Average: 15KMPH

Answer 9

accelleration = change in distance / change in time
so 4.17 m/s

Answer 10

kelsarhu maybe correct.