Answer
(a) The equation of the straight line that best fit the data is $a(t)=6.03\frac{m}{s^2}-0.510\frac{m}{s^3}t$.
(b) $v_{x}(t)=-0.255\frac{m}{s^3}t^2+6.03\frac{m}{s^2}t$
No, the graph is not a straight line
(c) The speed of the car at $t=5.00s$ is $23.8\frac{m}{s}$
(d) The distance the car travels between $t=0$ and $t=5.00s$ is $64.8m$
Work Step by Step
We take the initial position of the car to be the origin of our system.
From the given data, we see that the acceleration is not constant, so we cannot use the constant-acceleration kinematics equation.
(a) From the graph, the slope of the line is
$m=\frac{3.40-5.95}{5.00-0}=-0.510\frac{m}{s^3}$
Now, the equation of the straight line that best fit the data is given by
$a(t)=5.52\frac{m}{s^2}-0.510\frac{m}{s^3}(t-1.00s)$
$a(t)=6.03\frac{m}{s^2}-0.510\frac{m}{s^3}t$
(b) We know that the instantaneous acceleration is given by
$a=\frac{d}{dt}v_{x}$
Integrating the equation for acceleration from time $0$ to $t$, we get
$v_{x}(t)=v_{0x}-0.255\frac{m}{s^3}t^2+6.03\frac{m}{s^2}t$
$v_{x}(t)=-0.255\frac{m}{s^3}t^2+6.03\frac{m}{s^2}t$ , [$v_{0x}=0$]
The graph of $v$ vs $t$ is not a straight line.
(c) The speed of the car at $t=5.00s$ is
$v_{x}(5.00)=-0.255\frac{m}{s^3}(5.00s)^2+6.03\frac{m}{s^2}(5.00s)=23.8\frac{m}{s}$
(d) We know the instantneous velocity is given by
$v_{x}=\frac{d}{dt}x$
Integrating the equation for velocity from time $0s$ to $5.00s$, we get
$x(t)=x_{0}-\frac{0.255}{3}\frac{m}{s^3}(5.00s)^3+3.015\frac{m}{s^2}(5.00s)^2$
$x(t)=64.8m$ , [$x_{0}=0$]
Thus, the distance the car travels between $t=0$ and $t=5.00s$ is $64.8m$
