A! Good luck on your test!
Answer:
210 m
Explanation:
Given:
a = -6.8 m/s²
v₀ = 54 m/s
v = 0 m/s
Find: Δx
v² = v₀² + 2aΔx
(0 m/s)² = (54 m/s)² + 2 (-6.8 m/s²) Δx
Δx ≈ 210 m
Answer:
2.5 x 10⁷ J
Explanation:
F = thrust of the engine = 2.3 x 10⁵ N
d = distance traveled = 87 m
Work done by the engine is given as
W = F d = (2.3 x 10⁵) (87) = 200.1 x 10⁵ J
W' = Net work done
W'' = work done by catapult
KE₀ = initial kinetic energy = 0 J
KE = final kinetic energy = 4.5 x 10⁷ J
Net work done is given as
W' = KE - KE₀
W' = 4.5 x 10⁷ J
We know that
W' = W + W''
4.5 x 10⁷ = 2.001 x 10⁷ + W''
W'' = 2.5 x 10⁷ J
Answer:
4.408 m/s, 4.102 m/s, 4.026 m/s
Explanation:
The question is incomplete. The text of the original question states:
A race car moves such that its position fits the relationship
:

where x is measured in meters and t in seconds. Determine the instantaneous velocity of the car at t = 4.7 s, using time intervals of 0.40 s, 0.20 s, and 0.10 s.
We can find the instantanoues velocity of the car at any time t by calculating the derivative of the position, so we find:

And now we just need to substitute t=0.40 s, 0.20 s, and 0.10 s to find the corresponding instantaneous velocity:
