91Ó°ÊÓ

The car starts from the rest with an acceleration 4m/s²

At point A, the initial velocity v_i= 0 m/sec.

The acceleration gained by the car in time t₁= 6.0 sec is a₁= 4 m/s².

The car accelerates at the same rate till point B.

Thereafter, it travels with constant velocity say v₁, for time t₂=2.0 sec till the point C. The acceleration a₂is zero during this time.

After reaching point C, it starts deaccelerating with a₃= 3m/s².

The car stops at point D with final velocity v_f=0.

(1) Consider the first case when the car starts from starting point A
Here, the initial velocity of the car is $v_i=0m/sec$
The acceleration of the car is $a=4m/s^2$

The car travels with this acceleration for time $t_1=6.0sec$

Write the equation of motion to obtain the distance traveled by car during this time.

$d_1=v_i\left(t_1\right)+\frac{1}{2}{a}_1{\left(t_1\right)}^2$

Substitute the known variables in the above expression
$d_1=0+\frac{1}{2}\left(4\right){\left(6\right)}^2{d}_1=72m$
Thus, the distance traveled by car from point A to B is 72m
The velocity of the car during this time,

$v_1=v_i+a_1{t}_1$
Substitute the values,
$$\begin{gathered} v_1=0+4\left(6\right) \\ {v}_1=24m/s \end{gathered}$$

Thus, the car travels at the speed of 24 m/sec at this point

At point B, the velocity of the car is $v_1=24m/s$
The acceleration of the car is $a_2=0$
The time travelled is $t_2=2.0sec$
Write the equation of motion to determine the distance travelled by car from B to C

$$\begin{gathered} d_2=v_1{t}_2+\frac{1}{2}{a}_2{\left(t_2\right)}^2 \\ {d}_2=24\left(2\right)+0 \\ {d}_2=48m \end{gathered}$$$$\begin{gathered} d_2=v_1{t}_2+\frac{1}{2}{a}_2{t}_2 \\ {d}_2=24\left(2\right)+\frac{1}{2}\left(0\right)\left(2\right) \\ {d}_2=48m \end{gathered}$$

The distance travelled by car from point B to C is 48 m

The car travels at the same velocity $v_1=24m/sec$
The final velocity of the car is $v_f=0$
The acceleration of the car is $a_3=-3m/s^2$
Write the equation of motion to determine the time taken$t_3$ by car to travel from points C and D
$v_f=v_1+a_3{t}_3$
Substitute the values,
$$\begin{gathered} 0=24+(-3)t_3 \\ {t}_3=\frac{24}{3}=8sec \end{gathered}$$
Thus, the time taken by the car to stop is 8 sec
Now, the distance traveled by car during this time,

localid="1648119775304" $$\begin{gathered} d_3=v_1{t}_3+\frac{1}{2}{a}_3{t_3}^2 \\ {d}_3=24\left(8\right)+\frac{1}{2}(-3){\left(8\right)}^2 \\ {d}_3=96m \end{gathered}$$
Therefore, the distance traveled by car from point C to D is 96 m.

Now add all the distances, 72m + 48m + 96m = 216m
Hence, the distance between the two stop signs is 216 m