You are operating a radio-controlled model car on a vacant tennis court. Your position is the origin oc coordinates, and the surface of the court lies in the xy-plane. The car, which we represent as a point, has x and y-coordinates that vary with time according to x = 2.0 m – (0.25 m/s2)t2 y = (1.0 m/s)t + (0.025 m/s3)t3 1. Find the car’s coordinate’s and distance from you at time t = 2.0 s. A) (1.0m, 2.2 m) r = 2.4 m B) (2.2m, 1.5m) r = 2.7m C) (2.5m, 2.05m) r = 3.2 m D) (2.3m, 1.0m) r = 2.5m 2. The components of the car’s instantaneous velocity are the time derivatives of the coordinates: vx = dx/dt = (-0.25 m/s2)(2t), vy = dy/dt = 1.0 m/s + (0.025 m/s3)(3t2) This can be expressed as velocity vector v = vxî + vyĵ = (-0.5 m/s2)tî + [1.0 m/s + (0.075 m/s3)t2]ĵ 3. What are the components of instantaneous velocity and its magnitude at time t =2.0 s? A) vx = -1.0 m/s, vy = 1.3 m/s, v = 1.6 m/s, α128° B) vx = 2.3 m/s, vy = 1.5 m/s, v = 2.7 m/s, α128° C) vx = 1.0 m/s, vy = - 1.3 m/s, v = 1.6 m/s, α128° D) vx = -1.5 m/s, vy = 2.3 m/s, v = 2.7 m/s, α128° 4. Using the components of instantaneous velocity at any time t vx = dx/dt = (-0.25 m/s2)(2t), vy = dy/dt = 1.0 m/s + (0.025 m/s3)(3t2) Find the components of the average acceleration in the interval from t = 0.0 s to t = 2.0 s. A) avx = 0.5 m/s2, avy = - 0.015 m/s2 B) avx = - 0.5 m/s2, avy = 0.75 m/s2 C) avx = 0.15 m/s2, avy = 0.05 m/s2 D) avx = - 0.5 m/s2, avy = 0.15 m/s2
You are operating a radio-controlled model car on a vacant tennis court. Your position is the origin oc coordinates, and the surface of the court lies in the xy-plane. The car, which we represent as a point, has x and y-coordinates that vary with time according to
x = 2.0 m – (0.25 m/s2)t2
y = (1.0 m/s)t + (0.025 m/s3)t3
1. Find the car’s coordinate’s and distance from you at time t = 2.0 s.
A) (1.0m, 2.2 m) r = 2.4 m
B) (2.2m, 1.5m) r = 2.7m
C) (2.5m, 2.05m) r = 3.2 m
D) (2.3m, 1.0m) r = 2.5m
2. The components of the car’s instantaneous velocity are the time derivatives of the coordinates:
vx = dx/dt = (-0.25 m/s2)(2t),
vy = dy/dt = 1.0 m/s + (0.025 m/s3)(3t2)
This can be expressed as velocity vector
v = vxî + vyĵ = (-0.5 m/s2)tî + [1.0 m/s + (0.075 m/s3)t2]ĵ
3. What are the components of instantaneous velocity and its magnitude at time t =2.0 s?
A) vx = -1.0 m/s, vy = 1.3 m/s, v = 1.6 m/s, α128°
B) vx = 2.3 m/s, vy = 1.5 m/s, v = 2.7 m/s, α128°
C) vx = 1.0 m/s, vy = - 1.3 m/s, v = 1.6 m/s, α128°
D) vx = -1.5 m/s, vy = 2.3 m/s, v = 2.7 m/s, α128°
4. Using the components of instantaneous velocity at any time t vx = dx/dt = (-0.25 m/s2)(2t), vy = dy/dt = 1.0 m/s + (0.025 m/s3)(3t2)
Find the components of the average acceleration in the interval from t = 0.0 s to t = 2.0 s.
A) avx = 0.5 m/s2, avy = - 0.015 m/s2
B) avx = - 0.5 m/s2, avy = 0.75 m/s2
C) avx = 0.15 m/s2, avy = 0.05 m/s2
D) avx = - 0.5 m/s2, avy = 0.15 m/s2
5. What is the instantaneous acceleration at t = 2.0 s?
A)a = 0.75 m/s2, 149° C)a = 0.85 m/s2, 149°
B)a = 0.58 m/s2, 149° D)a = 0.30 m/s2, 149°
6. What is the strongest fundamental force?
A) weak nuclear force
C)
B) gravitational Force
D) strong nuclear force
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