Respuesta :
The acceleration of the car is given by:
[tex]a=\frac{v_f-v_i}{t}[/tex]
where
vi=16.0 m/s is the initial velocity of the car
vf=6.0 m/s is the final velocity of the car
t=4.0 s is the time interval
Substituting these data into the formula, we can find the car's acceleration:
[tex]a=\frac{6.0 m/s-16.0 m/s}{4.0 s}=-2.5 m/s^2[/tex]
and the negative sign is due to the fact the car is slowing down, so it is a negative acceleration.
The acceleration of the car during [tex]4.0 s[/tex] is [tex]\boxed{-2.5\,{\text{m/}}{{\text{s}}^{\text{2}}}}[/tex] i.e. towards the west.
Further explanation:
Acceleration is caused due to the force exerted on the body. In case when body is in equilibrium, the net force on the body will be zero therefore, acceleration will also be zero.
In case when body is moving with certain initial velocity and the net force on the body is zero, body will continue to move with the same velocity or we can say body is moving with uniform velocity.
Given:
Initial velocity of the car is [tex]16\text{ m/s}[/tex] towards east.
Final velocity of the car is [tex]6\text{ m/s}[/tex] towards east.
Time interval is .[tex]4.0\text{ s}[/tex].
Concept:
Instantaneous acceleration is defined as the rate of change of velocity with respect to time. Average acceleration is defined as the change of velocity over a certain period of time.
Mathematically,
[tex]\boxed{a = \dfrac{{\Delta V}}{{\Delta t}}}[/tex] ……(1)
Here, [tex]a[/tex] is the average acceleration of car, [tex]\Delta V[/tex] is the change in velocity and [tex]\Delta t[/tex] is the time interval.
Write the expression for change in velocity
[tex]\Delta V = {V_{\text{f}}} - {V_{\text{i}}}[/tex] …… (2)
Here, [tex]{V_{\text{f}}}[/tex] is the final velocity of car and [tex]{V_{\text{i}}}[/tex] is the initial velocity of the car.
Velocity is a vector quantity so take magnitude as well as direction of the velocity, while substituting in the above expression.
Let’s assume unit vector in the east direction is [tex]\hat i[/tex] and unit vector in west direction is [tex]- \hat j[/tex].
Substitute [tex]6\,\hat i[/tex] for [tex]{V_{\text{f}}}[/tex] and [tex]16\,\hat i[/tex] for [tex]{V_{\text{i}}}[/tex] in equation (2).
[tex]\begin{aligned}\Delta V&=\left( {6{\kern 1pt} \hat i} \right) - \left( {16\,\hat i\,} \right)\\&=10\,\left( { - \hat i} \right)\\\end{aligned}[/tex]
Substitute [tex]10\,\left( { - \hat i} \right)[/tex] for [tex]\Delta V[/tex] and [tex]4.0\text{ s}[/tex] for [tex]\Delta t[/tex] in equation (1).
[tex]\begin{aligned}a&=\frac{{\left( { - 10\,\hat i\,{\text{m/s}}} \right)}}{{\left( {4.0\,{\text{s}}} \right)}} \\&=2.5\,\left( { - \hat i} \right)\,{\text{m/}}{{\text{s}}^{\text{2}}} \\ \end{aligned}[/tex]
Thus, the acceleration of the car during [tex]4.0 s[/tex] is [tex]\boxed{-2.5\,{\text{m/}}{{\text{s}}^{\text{2}}}}[/tex] i.e. towards the west.
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Answer Details:
Grade: High School
Subject: Physics
Chapter: Kinematics
Keywords:
Car, initial, velocity, 16.0 m/s, east, 16 m/s, 16 meter per second, 16.0 meter per second, uniformly, 6.0 m/s, 6 m/s, 6.0 meter per second, 6 meter per second, 4.0 second, 4.0 s, 4 second, 4 s, acceleration, during, interval.
