use consevation of linear momentum
- m1v1+m2v2=(M1+M2)V3
- 281(2.82)+209(-1.72)=(209+281)V2
- 792.42-359.48=490v3
- 432.9=490v3
- V3=0.88m/s
Data:
u=0 m/s is the initial velocity of the plane
v=62 m/s is the final velocity of the plane (at which the plane takes off)
a=1.7 m/s^2 is the acceleration of the plane
To find the minimum distance S the plane needs to take off, we can use the following equation:
Re-arranging it and substituting the numbers, we find
<span>speed of slower=x; distance in 6 hours is 6x
speed of larger=2x; distance in 6 hours is 12x
12x-6x=204 miles
</span><span> 6x=204
x=34 mph slower, 204 miles in 6 hours.
2x=68 mph, faster, 408 miles in 6 hours, and difference is 204 miles</span>
Answer:
A) 3.13 m/s
B) 5.34 N
C) W = 26.9 J
Explanation:
We are told that the position as a function of time is given by;
x(t) = αt² + βt³
Where;
α = 0.210 m/s² and β = 2.04×10^(−2) m/s³ = 0.0204 m/s³
Thus;
x(t) = 0.21t² + 0.0204t³
A) Velocity is gotten from the derivative of the displacement.
Thus;
v(t) = x'(t) = 2(0.21t) + 3(0.0204t²)
v(t) = 0.42t + 0.0612t²
v(4.5) = 0.42(4.5) + 0.0612(4.5)²
v(4.5) = 3.1293 m/s ≈ 3.13 m/s
B) acceleration is gotten from the derivative of the velocity
a(t) = v'(t) = 0.42 + 2(0.0612t)
a(4.5) = 0.42 + 2(0.0612 × 4.5)
a(4.5) = 0.9708 m/s²
Force = ma = 5.5 × 0.9708
F = 5.3394 N ≈ 5.34 N
C) Since no friction, work done is kinetic energy.
Thus;
W = ½mv²
W = ½ × 5.5 × 3.1293²
W = 26.9 J
Answer:
accelerated
Explanation:
The motion of the body where the acceleration is constant is known as uniformly accelerated motion. The value of the acceleration does not change with the function of time.
