Cars A and B move in the same direction in adjacent lanes. The position x of car

Question

Cars A and B move in the same direction in adjacent lanes. The position x of car A is given in the figure, from time t = 0 to t = 7.0 s. The figure's vertical scaling is set by x_s = 16.0 m. At t = 0, car B is at x = 0, with a velocity of 6.00 m/s and a negative constant acceleration a_B. What must a_B be such that the cars are (momentarily) side by side (momentarily at the same value of x) at t = 4.0 s? For that value of a_B, how many times are the cars side by side? How many times will the cars be side by side if the magnitude of acceleration a_B is more than and less than the answer to part (a)? Number _______ Units Number _______ Units Number _______ Units Number _______ Units

Explanation / Answer

constant at v = 2m/s. For this car, then, x(t) = 16m + 2m/s * t
For the second car, we have s(t) = 4m/s * t + ½*aB*t².

a) at t=4, x(16) = 16m + 32m = 48m
and s(4)

48m = 24m + ½*aB*16m

48m = 24m + 8*aB m
Then aB = 24/8 = 3 m/s²


b) s(t) = 6t + 2t² = x(t) = 16 + 2t
2t2 + 4t - 16 = 0
t= 2s, so "cars are side by side once".

c) cars be side by side never

d) cars be side by side Twice.

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