JOsiNg Re answers to parts (a) and (b) as the initial velocity and position for

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JOsiNg Re answers to parts (a) and (b) as the initial velocity and position for the motion of the car during braking, what total distance does the car travel? Noed Help? Rand Traumatic brain injury such as concussion results when the head undergoes a very large time will not cause injury, whereas an acceleration greater than 1,000 m/s2 lasting for at least 1 ms will cause injury. Suppose a small child rolls off a bed that is 0.39 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 2.1 mm. If the floor is carpeted, this stopping distance is increased to about 1.4 cm. Calculate the magnitude and duration of the deceleration in both cases, to determine the risk of injury. Assume the child remains horizontal during the fall to the floor. Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate. hardwood floor magnitude hardwood floor duration carpeted floor magnitude acceleration. Generally, an acceleration less than 800 m/s2 lasting for any length of ms m/s 1.5 carpeted floor duration differs from the correct answer by more than 10%. Double check your calculations. ms Need Help? RendIt Notes O Ask Your Teach A tennis player tosses a tennis ball straight up and then catches it after 2.21 s at the same height as the point of release. acBook A

Explanation / Answer

Velocity gained during drop from a height of 0.39 m can be calculated by the equation v2 = u2 + 2as

u (initial velocity) = 0, a (acceleration = 9.8 m/s2), and s (distance) = 0.39 m

v = (2as) =(2×9.8×0.39) = 2.765 m/s

Magnitude of deceleration and time can again be calculated by the equations v2 = u2 + 2as and v = u + at

For hardwood floor, v (final velocity) = 0, u (initial velocity) = 2.765, and s (distance) = 2.1 mm

a = (v2 - u2)/2s = -1820 m/s2;

Using the above value of a, t = (v - u)/a = (0 - 2.765)/(-1820) = 0.0015 s = 1.5 ms

Similarly, for carpet floor, v (final velocity) = 0, u (initial velocity) = 2.765, and s (distance) = 1.4 cm

Again, a = (v2 - u2)/2s = -273 m/s2;

Using the above value of a, t = (v - u)/a = (0 - 2.765)/(-273) = 0.01 s = 100 ms

[Please note that a negative value of acceleration implies deceleration and the negative (minus) sign can be ignored; moreover, the question is only asking the 'magnitude'.]

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