Question

i have a velocity of +9.1 m/s. I need the acceleration in m/s2? how do i find this with such little information?

Answer 1

Is that really all you have?

Answer 2

the question reads; NASA has developed Deep-Space (D-1), a spacecraft that is scheduled to rendezvous with asteroid named 1992 KD (which orbits the sun millions of miles from the earth). The propulsion system of DS-1 works by ejecting high speed argon ions out the rear of the engine. The engine slowly increases the velocity of D-1 by about +9.1 m/s per day. What is the acceleration of D-1 in m/s2?

Answer 3

Wow I've never seen a question like this before....seems impossible when you look at it but I'm just throwing things out there so try this

As to (a), V = U + AT = 0 + 9.72 m/s/d * T = 2630 m/s

So T = 2630 m/s // 9.72 m/s/d = 2630/9.72 = 270.5761317 days. ANS if it's not it then I have no idea what it could be

Answer 4

now I'm lost as to where you got all those numbers..

Answer 5

you can follow this example and hopefully you'll get the answer...hope this helps
19m/s/day = 19 m/s/(24 hr/day*60min/hr*60sec/min*1 day) = 2.2×10^-4 m/s^2

Answer 6

acceleration is the change in velocity over time. so\[a=\frac{\Delta v}{t}\] which is what ruthym is doing in the comment above mine.

Answer 7

Using what @zephyr141 has noted (I think @ruthym transposed the 9 and 1 and ended up using 19 m/s/day), and doing the math, you get:
(9.1)/(24*60*60)=0.00010532407
delta v (change in velocity - in m/s - over the time of a full day) = 9.1 m/s
total time in seconds (over a day) for delta v = 24*60*60 ... (24 hrs x 60 min/hr x 60 sec/min = 86,400)
Acceleration in m/s^2 (or m/s/s) is 0.00010532407 (or 1.0532407e-4) m/s^2

To verify that 1.0532407e-4 m/s^2 is the correct acceleration in m/s^2, copy and paste the following values for the equation a * t :
(1.0532407e-4)*(24*60*60)=
into the Google One Line (the single line where you enter information you want Google to search for), and then hit enter, the answer of 9.099999648 will appear in the Google calculator. Pretty close to 9.1 I think ;)

So, even though the velocity was given in m/s, the change in that velocity occurred over a full day. When using v/t for solving for a, t has to be expressed in total seconds that acceleration took place over to result in a velocity increase of 9.1 m/s (the units of time, as well as any other units, must be the same).