91影视

Initial velocity is $55\text{km/h}$

Acceleration (a) is $5.18{\text{m/s}}^{\text{2}}$

Distance $x_1=40\text{m}$, Time $T_1=2.8\text{s}$

Distance $x_2=32\text{m}$, Time $T_2=1.8\text{s}$

The problem deals with the kinematic equation of motion in which the motion of an object is described at constant acceleration. Using the formula for the distance and velocity, the distance during reaction time, breaking time and hence total distance can be calculated. Using the calculated quantities it can be determine whether driver should stop or continue to move at $\mathbf{55}\mathbf{}\mathbf{km}\mathbf{/}\mathbf{h}$at different distances and duration.

Formula:

The velocity in kinematic equation is given by,

$v_f^2=v_0^2+2ax$

Let鈥檚 consider the direction of motion is +x.

so$a=-5.18{\text{m/s}}^{\text{2}}$ and

$\begin{array}{rcl}{v}_0& =& 55\text{km/h}\\ & =& 15.28\text{m/s}\end{array}$

The distance traveled during the reaction time is

$\begin{array}{rcl}{d}_r& =& v_{0}T\\ d_r& =& \left(15.28\frac{\text{m}}{\text{s}}\right)\left(0.75\right)\\ d_r& =& 11.46\text{m}\end{array}$

The distance traveled during the breaking is

$\begin{array}{rcl}{v}^2& =& {v_0}^2+2a{d}_b\\ d_b& =& -\frac{15.28{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}}{2\left(-5.18{\frac{\text{m}}{{\text{s}}^{\text{2}}}}^2\right)}\\ & =& 22.53\text{m}\end{array}$

Thus, total distance traveled is

$\begin{array}{rcl}{d}_r+d_b& =& 11.46\text{m}+22.53\text{m}\\ & =& 34.0\text{m}\end{array}$

So driver is able to stop in time. If he continues at$v_0$, the car would enter the intersection

$\begin{array}{rcl}t& =& \frac{40\text{m}}{5.28\text{m/s}}\\ & =& 2.6\text{s}\end{array}$

This is sufficient to enter the intersection before light turns. But it is too small to continue driving. So it would be advisable for the driver to stop.

Total distance to stop is greater than the distance to the intersection; hence the driver can鈥檛 stop without the front of the car being few meters into the consideration.

Time to reach this distance is$\frac{32}{15.28}\text{m/s}=2.1\text{s}$.

This time is greater than that of the available time. So this is tricky situation. Driver would not be able to stop and would be able to just cross the intersection.