Can you drive safe ??
On a foggy day, the range of visibility is
.
If your vehicle has deceleration(braking) power of
, what is the max. speed with which you can drive safe??
Details and assumptions:
Take reaction time =
Give your answer in m/s.
Round off your answer to nearest natural no.
Driving safe means that you don't hit anyone with your vehicle.
The answer is 13.
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Initially, I'd be driving my car at a constant speed of u m/s. Upon seeing an obstacle 10m ahead of me, I'd cover a distance of dm for 0.1 s before commencing deceleration. So u= d/0.1= 10d.
At this point, the distance between the car and the obstacle, s, is equal to (10-d)m and my speed is 10d. The speed at which the car travels such that it hits the obstacle can be determined using the linear motion equation v²=u²+2as, where v is the final speed, u is the initial speed, a is the acceleration and s is the distance covered.
(10d)²+2(-10)(10-d)= 0 100d²+20d-200= 0
Solving the inequality above, we get d= [(-1+√201)/10], [(-1-√201)/10]. However, we only consider the first root since the second root is negative and the distance covered during deceleration is positive.
Therefore, d= [(-1+√201)/10]=1.317744688.
The corresponding initial speed= (10)(1.317744688)=13.17744688 m/s= 13 m/s to the nearest natural number. I would have to drive at a speed lower than this to avoid hitting the obstacle, thereby rendering me a safe driver.