An airplane starts from rest and accelerates at 12.1 m/s^2. What is its speed at

Question

An airplane starts from rest and accelerates at 12.1 m/s^2. What is its speed at the end of a 500 m nmway? A) 98m/s B) 110m/s C) 60.5 m/s D) 127 m/s E) 500 m/s 7. The position of a particle moving along the x-axis is given by x - 30 t^2 - 30 t^4, where x is in meters and t is in seconds. What is the position of the particle when it achieves its maximum speed in the positive x-direction? A) 5.6 m B) 9.8 m C) 4.2 m D) 27 m E) 3 m 8. When a drag strip vehicle reaches a velocity of 60 m/s, it begins a negative acceleration by releasing a drag chute and applying its brakes. While reducing its velocity back to zero, its acceleration along a straight line path is a constant -9.5 m/s^2,. What displacement does it undergo during this deceleration period? A) 40 m B) 320 m C) 77 m D) 240 m E) 189 m 9. A car slows down from a speed of 31 m/s to a speed of 12 m s over a distance of 380 m. How long does this take, assuming constant acceleration? A) 25 sec B) 17.7 sec C) 21.9 sec D) 34.2 sec E) 14.6 sec

Explanation / Answer

6) using vf^2 - vi^2 = 2ad

v^2 - 0 = 2(12.1)(500)

v = 110 m/s

Ans(B)

7. v = dx/dt = d(30t^2 - 30t^4) / dt

v = 60t - 120t^3

v will be maximum or minimum when dv/dt = 0

dv/dt = 60 - 360t^2 = 0

t^2 = 60/360 = 1/6

t = 0.408 s

position at 0.408 s ..

x = 30 x 0.408^2 - 30 x 0.408^4 = 4.16 m

Ans(c)

8. using vf^2 - vi^2 = 2ad

0^2 - 60^2 = 2(-9.5)(d)

d = 189.4 m

Ans(E)

9. vf^2 - vi^2 = 2ad

31^2 - 12^2 = 2(a)(380)

a = 1.075 m/s^2 ]

now using , v = u + at

31 = 12 + 1.075t

t = 17.67 s

Ans(B)

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