My Calculating Acceleration & Distance of a Car Moving at 70 km/h

[BigCountry]

A car moving at 70 lm/h accelerates to a speed of 110 km/h in a time of 5 secs.

a) What was the cars acceleration?

b) How far did the car travel while accelerating for the 5 secs?

My Ans for a):

V1 (initial velocity)=70 km/h or 7000 m/s
v2 (final velocity)=110 km/h or 110000 m/s
t (time)=5 secs
Find a (acceleration)

a = V2 - V1 / t2 - t1
a = 110 - 70 / 5 - 0
a = 40 / 5
a = 8 km/h or 8000 m/s

I am not even close and believe I am wrong in converting to consistent units also. Am I even using the right formula? As for ans b), I believe I need to figure out the acceleration first. When I have that, I think the formula I should be using to figure out b) is d = (V1 + V2) / 2 x t. Is this even correct?

Your help is much appreciated,

 

[civil_dude]

you converted the km to meters, but you didn't convert the 1/hr to 1/seconds. i.e. 70km/h does NOT = 7000 m/s, it = 116.7 m/s.

 

[Päällikkö]

Your conversions from km/h to m/s are wrong. The correct ratio is 1/3.6 rather than 1000.
You've confused the units: You will not get units km/h² nor m/s² with (km/h)/s.

 

[BigCountry]

My UPDATED Ans for a):

V1 (initial velocity)=70 km/h or 70000 m/h
v2 (final velocity)=110 km/h or 110000 m/h
t (time)=5 secs x 3600 secs/hr = 18000
Find a (acceleration)

a = V2 - V1 / t2 - t1
a = 110000 - 70000 / 18000 - 0
a = 40000/ 18000
a = 2.2 m/s

Thanks for your help on this one guys. Now b).

A car moving at 70 lm/h accelerates to a speed of 110 km/h in a time of 5 secs.

b) How far did the car travel while accelerating for the 5 secs?

The formula I used is: d = (V1 + V2) / 2 x t

d = (70000+110000) / 2 x 18000
d = 180000 / 2 x 18000
d = 90000 x 18000
d = 1.62 09 ?

I am obviously using the wrong formula. Please help.

 

[BobG]

First off, you're usually better off converting to meters and seconds. It doesn't matter too much on your first problem, since you wound up with the same ratio. Still...

70 km/hr = 19.44 m/s
110 km/hr = 30.56 m/s

You still wind up with an acceleration of 2.2 m/sec^2 (not 2.2 m/s)

For the second problem, at least remember your basic equation for distance:

s_f=s_i + v_it + \frac{1}{2}at^2

(Or, you can be British and use 's' for final position, 'u' for initial position, and you wind up with an easy to remember acronym: suvat)

You can rearrange this for several different short cuts, but this equation works just about all of the time.

 

[BigCountry]

My Ans for b)

d = Vi t + at^2/2
d = 19.44*5 + 2.2(5^2)/2
d = 97.2 + 27.5
d = 124.7 m

Thanks for all of your help,
Much, much, much appreciated!

 

[lightgrav]

You should recognize that "5 seconds" is NOT 18000 hours of time ...

You knew that much of the difficulty of this problem was going to be UNITS,
yet you refused to use them for the time conversion [18000 SEC^2 / HOUR]
even after you screwed up the speed conversion!

By the way, YOUR way of finding the distance traveled is better than BobG's , since yours does not use the (possibly wrong) computed result.