I really need help with this problem?

Question

A body moving with uniform acceleration has a velocity of 15.5 cm/s when its x coordinate is 3.59 cm. If its x coordinate 1.87 seconds later is -8 cm, what is the magnitude of its acceleration?

Can you show your work too? Thanks.

cut my wrists :(

Anyway, how can you find the average velocity when the distance is unknown.

Rawr.

Answer 1

At t = 0:
x(0) = 3.59 cm
v(0) = 15.5 cm/s

At t = 1.87:
x(1.87) = - 8 cm

a) We are told that it is undergoing constant acceleration. Therefore,
v(t) = v(0) + a*t

If you integrate that over time, you get:
x(t) - x(0) = v(0)*t + (a*t^2)/2, or
x(t) = x(0) + v(0)*t + (a*t^2)/2

But we know x(0), v(0) and x(1.87), so we can solve this:
x(1.87) - x(0) - v(0)*1.87 = a*(1.87^2)/2

a = 2*(x(1.87) - x(0) - v(0)*1.87)/(1.87^2)
= 2*(-8 - 3.59 - 15.5*1.87)/(1.87^2)
= -23.2 (cm/s^2)

(Spreadsheets are great for doing this sort of computation.)

Answer 2

I won't do it for you, but I'll tell you how.
You've got an initial position and a velocity.
You've got a time difference and a second position.
That gives you an average velocity.
The acceleration is twice the velocity difference.