Respuesta :
Answer:
[tex]4.08cm/s^2[/tex]
Explanation:
The second equation of a uniformly accelerated motion could be used to solve this problem. This is given by equation (1);
[tex]s=ut+\frac{1}{2}at^2....................(1)[/tex]
where u is the particle's initial velocity, t is the time taken, a is the acceleration and s is the distance travelled.
Given;
u = 20cm/s
t = 7s
a = ?
s = ?
The particle moved from one point [tex]x_1[/tex] to another point [tex]x_2[/tex] along the x-axis, where [tex]x_1=10cm[/tex] and [tex]x_2=-30cm[/tex]. This information could be used to find the distance s covered by the object as follows;
[tex]s=x_1-x_2.................(2)\\s=10-(-30)\\s=10+30\\s=40cm[/tex]
We the make appropriate substitutions into equation (1) and then solve for the acceleration.
[tex]40=(20*7)+\frac{1}{2}*a*7^2\\40=140+\frac{1}{2}*a*49\\40=140+24.5a\\40-140=24.5a\\hence\\24.5a=-100\\a=\frac{-100}{24.5}\\a=-4.08cm/s^2[/tex]
The negative sign is an indication that the particle is decelerating.
Answer:
7.347 cm / s²
Explanation:
Using equation of linear motion
S = ut + 1/2 at²
where total displacement = final displacement - initial displacement
S = - 30 - 10 = - 40 cm
- 40 cm = (20 cm /s × 7 s) + 1/2 a (7²)
- 40 cm = 140 cm + 1/2 49 a
- 40 cm - 140 cm = 1/2 × 49 a
- 180 cm × 2 / 49 s² = a
a = -7.347 cm / s²
It is probably decelerating.
