can someone please help? Cars A and B arc (raveling at vA = 72 mph and vB = 67 m

Question

can someone please help?

Cars A and B arc (raveling at vA = 72 mph and vB = 67 mph, respectively, when the driver of car B applies the brakes abruptly, causing the car to slide to a stop. The driver of car A takes 1.5 s to react to the situation and applies the brakes in turn, causing car A to slide as well. If A and B slide with equal accelerations, i.e.. suml A = suml B = - mu kg, where mu k = 0.83 is the kinetic friction coefficient and g is the acceleration of gravity, compute the minimum distance d between A and B at the time B starts sliding to avoid a collision.

Explanation / Answer

first let us calculate the distace covered by the car B during this whole incident

given intial velocity of B = 67 miles / hour = 29.95 m/s

final velocity = 0 as it will come to rest ...

acceleration =-Mu*g = -0.83*9.81 =-8.14 m/s^2

so the distance covered by the car = v^2 - u^2 / (2*acceleration) = 0^2 - (29.95)^2 / (2*-8.14) = 55.08 meters

now the car A

let D be the distance between the cars intially

the car must travel less than D + 55.08 meters to prevent the accident

inital velocity =72 mph = 32.19 m/s

but given that the driver takes 1.5 seconds to respond

so the distance travelled during this time = 1.5* 32.19 = 48.285 meters

so the net distance = D + 55.08 -48.285 meters = D+6.975 meters

the distance covered by the car during decelrating must be less than or equal to  D+6.975 meters to prevent the accident

so

v^2 - u^2 / (2*acceleration) <= D+6.975

0^2 - (32.19)^2 / (2*-8.14) <= D+6.975

63.648 = D+6.975

so D => 56.673 meters to prevent the accident

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