Welcome to our community

Be a part of something great, join today!

Physics 2.3.16 constant acceleration

karush

Well-known member
Jan 31, 2012
2,724
2.3.16 A car is traveling at $45 \, km/h$ at time $t=0$ It accelerates at a constant rate of $10 \, km/h\, s$
(a) How fast is the care going at $t=1\, s$?
$$v_t=v_0+at=45+10(1)=55\,\dfrac{km}{h}$$
at $t=2\,s$
$$v_t=v_0+at=45+10(2)=45+20=60\,\dfrac{km}{h}$$
(b) What is its speed at a general time t




ok this is a very simple problem but when you have constant acceleration there is no power on s?
also (b) what is meant by general time t is that an average or an equation.


also typos perhaps...
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,123
2.3.16 A car is traveling at $45 \, km/h$ at time $t=0$ It accelerates at a constant rate of $10 \, km/h\, s$
(a) How fast is the care going at $t=1\, s$?
$$v_t=v_0+at=45+10(1)=55\,\dfrac{km}{h}$$
at $t=2\,s$
$$v_t=v_0+at=45+10(2)=45+20=60\,\dfrac{km}{h}$$
(b) What is its speed at a general time t




ok this is a very simple problem but when you have constant acceleration there is no power on s?
also (b) what is meant by general time t is that an average or an equation.


also typos perhaps...
First of all we need to fix that unit in the acceleration, which is a bit weird. Since a) is putting time in seconds, then let's get the acceleration to km/s^2:
\(\displaystyle \dfrac{10 ~\text{km}}{\text{h s}} \cdot \dfrac{10 ~ \text{h}}{3600 ~ \text{s}} = 0.278 ~ \text{km/}s^2\)

Now use \(\displaystyle v_0 + at\).

-Dan
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
I disagree with Dan. Since we want the speed, in km/h, after t seconds, an acceleration in "km/hs", kilometers per hour per second, is simple and perfectly reasonable.

Now, to the given question. (a) asked for the speed after 1 second at an acceleration of 10 km/hs. Yes, multiplying 10 km/hs
by 1 second
gives an increase of 10 km/h so the speed goes from 55 km/h to 65 km/h. I notice that you next give the speed after 2 seconds acceleration. That doesn't appear to have been asked but it was not a bad thing to do- you multiplied 10 km/hs by 2 seconds to get an increase of 20 km/h. What if, instead of "1 second" or "2 seconds" you were told the acceleration lasted for "t seconds". You would do exactly the same thing: multiply 10 km/hs by t seconds to get an increase in speed of 10t km/h.