Question

An object moving with uniform acceleration has a velocity of 12.0 cm/s in the positive X direction when its X coordinate is 3.00 cm. If its X coordinate 2.00 s later is -5.00 cm, what is its acceleration?

Answer 1

Hey.

Here is the answer.

Let acceleration be a .

t = 2 sec

s1= 3cm , s2= -5cm , u at s1 = 12cm/s in +ve X

We know that s = u t + 1/2 a t^2

so, s = s2-s1 = 12×2 + 1/2×a×2^2

or, -5-3 = 24 + 2a

or, -8 -24 = 2a

or, -32 = 2a

so, a = -16 cm/s

Thanks.

Answer 2

Answer: -16cm/s^2

Explanation: t=2s, s1=3cm, s2= -5cm, u at s1 = 12cm/s

Now, s=ut+1/2at^2

=> s2-s1= ut+1/2at^2

=> -5-3 = 12×2+1/2×a×2×2

=> -32 = 2a

=> -16m/s^2 = a