Physics: Finding one dimensional acceleration without being given the time?

Question

I'm new to physics and I'm trying to get a head start by doing some practice before school starts. I'm guessing this problem is really simple, but I keep getting the wrong answer.

I know that acceleration = velocity / time, but you aren't given any values of time in this problem.

Can someone please show me how to solve this step by step?

Problem:

"An object with constant acceleration has velocity v = 10 m/s when it is at x = 6 m and v = 15 m/s when it is at x = 10 m. What is its acceleration?"

Textbook says correct answer is 15.6 m/s^2

Answer 1

Try using the Kinematics equations

Answer 2

You may use one of the following equations to solve your problem:

v = u + at

s = ut + (1/2)a(t^2)

v^2 = u^2 + 2as

u - initial speed

v - final speed

a - acceleration

s - displacement

t - time

The solution to your problem is an application of the 3rd equation.

Assume time, t=0 when x=6m. Then,

v = 15 m/s

u = 10 m/s

s = 10-6 = 4m

Therefore, 225 = 100 + 2 * 4 * a

so ... a = (225-100)/8 = 125/8 = 15.625 m/s^2

Answer 3

Since you haven't been given an *explanation* proper, let me offer *that*.

If it constantly accelerates from 10 m/s to 15 m/s, then that means that its *average* velocity was 12.5 m/s.

But it traveled 4 meters with an average velocity of 12.5 m/s, then that means that it traveled for 4 m / (12.5 m/s) = 0.32 s.

That general strategy (dt = (x1 - x0) / [ (v1 + v0)/ 2 ] ) can be used to do constant-acceleration physics in general. So, for example, you get that:

a = (v1 - v0) / dt = (v1 - v0) (v1 + v0) / (2 (x1 - x0) )

But (a - b)*(a+b) = a² - b², so:

2 a (x1 - x0) = (v1)² - (v0)².

QED.

Answer 4

v^2 - u^2 = 2as

where v = final velocity

u = initial velocity

a = accelaration

s = distance travelled between velocities u & v

s =10 -6 = 4m

v = 15m/s

u = 10m/s

225 - 100 = 8 a

a = 125/8 = 15.625 m/sec^2

Answer 5

* hides behind desk * Ugh I hate Physics ..good luck.