Question

Suppose that a teacher driving a 1972 LeMans zooms out of a darkened tunnel at 29.3 m/s. He is momentarily blinded by the sunshine. When he recovers, he sees that he is fast overtaking a bus ahead in his lane moving at the slower speed of 14.3 m/s. He hits the brakes as fast as he can (his reaction time is 0.31 s). If he can decelerate at 3.3 m/s2, what is the minimum distance between the driver and the bus when he first sees it so that they do not collide?

Answer 1

Given,

Speed of the car $(u)=29.3 m/sec$

Speed of bus $(U)=14.3 m/sec$

Reaction time $(t_r)=0.31 sec$

Deceleration $(a)=-3.3 m/sec^2$

Let the time gap between the bus and car after seeing the bus is t,

Let the distance covered by bus in t time is $x$ and initial distance between bus and car is $D$

So, $x= U\times t = 14.3 t ------(i)$

distance covered by car in the same duration of time $D+x=u\times t+(u(t-0.31)-\dfrac{a(t-0.31)^2}{2}) -----(ii)$

From equation (i) and (ii),

Substituting the value of $x,$ in equation (ii)

$D+14.3t=u\times t+(u(t-0.31)-\dfrac{a(t-0.31)^2}{2})$

$\Rightarrow D=29.3\times 0.31 -14.3t+29.3t-29.3\times 0.31-\dfrac{3.3(t-0.31)^2}{2}$

$\Rightarrow D=15t-\dfrac{3.3(t-0.31)^2}{2}$

taking the differentiation of both side,

$\dfrac{dD}{dt}=15-3.3(t-0.31)$

for maxima and minima,

$\dfrac{dD}{dt}=0$

$\Rightarrow 15-3.3(t-0.31)=0$

$\Rightarrow 3.3(t-0.31)=15$

$\Rightarrow t-0.31=\dfrac{15}{3.3}$

$\Rightarrow t=(0.31+4.54) sec$

$\Rightarrow t =4.85 sec$

Hence, minimum distance between the car and bus

$\Rightarrow D=15\times 4.85- \dfrac{3.3(4.85-0.31)^2}{2}$

$\Rightarrow D =(72.75-33.41)m$

$\Rightarrow D=39.34 m$