Respuesta :
From the initial and final velocity, we can calculate the average acceleration of the object during its motion:
[tex]a= \frac{v_f - v_i}{t} = \frac{11.0 m/s-0 m/s}{5.0 s}=2.2 m/s^2 [/tex]
For an uniform accelerated motion, the law of motion for the position x(t) is (starting from x=0 and at rest, so v0=0):
[tex]x(t) = \frac{1}{2} a t^2[/tex]
we can put in the acceleration we found and the time to verify if this corresponds to the value given by the problem (x=40.0 m):
[tex]x(t)= \frac{1}{2}(2.2 m/s^2)(5.0 s)^2=27.5 m [/tex]
and this value does not corresponds to 40.0 m, so the object's acceleration was non-uniform.
[tex]a= \frac{v_f - v_i}{t} = \frac{11.0 m/s-0 m/s}{5.0 s}=2.2 m/s^2 [/tex]
For an uniform accelerated motion, the law of motion for the position x(t) is (starting from x=0 and at rest, so v0=0):
[tex]x(t) = \frac{1}{2} a t^2[/tex]
we can put in the acceleration we found and the time to verify if this corresponds to the value given by the problem (x=40.0 m):
[tex]x(t)= \frac{1}{2}(2.2 m/s^2)(5.0 s)^2=27.5 m [/tex]
and this value does not corresponds to 40.0 m, so the object's acceleration was non-uniform.
The object’s acceleration is non-uniform.
Further explain:
Here, we have to find whether the motion of the object is in uniform acceleration or non-uniform acceleration.
Initially, the distance travel by object is zero.
The velocity of the object is zero that is object is in rest position.
After time duration of [tex]5{\text{ s}}[/tex].
The object is at the distance of [tex]40{\text{ m}}[/tex] from its initial position.
The velocity of the object at that position is [tex]11{\text{ }}{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}[/tex].
If the acceleration is uniform then the calculated value of the distance travel by the object will be same as the given distance.
The uniform acceleration can be calculated as,
[tex]\boxed{v = u + at}[/tex] …… (1)
The distance travel by the object with uniform acceleration can be calculated as,
[tex]\boxed{s = ut + \frac{1}{2}a{t^2}}[/tex] …… (2)
For uniform acceleration,
Substitute [tex]11{\text{ }}{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}[/tex] for [tex]v[/tex] , [tex]0{\text{ }}{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}[/tex] for [tex]u[/tex] and [tex]5{\text{ s}}[/tex] for [tex]t[/tex] in equation (1).
[tex]\begin{aligned}11&=0+a\left(5\right)\\11&=5a\\a&=\frac{{11}}{5}\\a&=2.2{\text{ }}{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {{{\text{s}}^2}}}} \right. \kern-\nulldelimiterspace}{{{\text{s}}^2}}}\\\end{aligned}[/tex]
The distance travelled by the object within the time duration of [tex]5{\text{ s}}[/tex] is,
Substitute [tex]40{\text{ m}}[/tex] for [tex]s[/tex] , [tex]0{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}[/tex] for [tex]u[/tex] and [tex]5{\text{ s}}[/tex] for [tex]t[/tex] in equation (2).
[tex]\begin{aligned}s&=0\times 5 + \frac{1}{2} \times \left( {2.2} \right){\left( 5 \right)^2}\\&=1.1 \times 25\\&=27.5{\text{m}}\\\end{gathered}[/tex]
Here, the value of the distance travel by the object with uniform acceleration is different from the given distance. So, the motion of the object is non-uniform acceleration.
Learn more:
1. For the angular speed with coefficient of friction: https://brainly.com/question/9575487.
2. Velocity of ball when maximum caught it: https://brainly.com/question/11023695.
3. Motion under friction https://brainly.com/question/7031524.
Answer detail:
Grade: Senior school
Subject: Physics
Chapter: Kinematics
Keywords:
Object, five second later, object observed to be, acceleration uniform or non-uniform, motion, equation of motion, Newton's laws, distance, acceleration, acceleration due to gravity.
