Answer:
304.89m
Explanation:
Given
acceleration a = 2.52m/s²
final speed v = 39.2m/s
initial speed = 0m/s (car accelerates from rest)
Using the equation of motion below to get the distance of Doc brown from Marty;
v² = u²+2as
substitute the given parameters
39.2² = 0²+2(2.52)s
1536.64 = 0+5.04s
divide both sides by 5.04
1536.64/5.04 = 5.04s/5.04
rearrange the equation
5.04s/5.04 = 1536.64/5.04
s = 304.89m
Hence He and Marty must stand at 304.89m to allow the car to accelerate from rest to a speed of 39.2 m/s?
Upstream speed = S - 1
Downstream speed = S + 1
Average speed = total distance / total time
Average speed = (S - 1) + (S + 1) / 2
= S
S = 6 miles / 4 hours
S = 1.5 miles per hour
Answer:
1.53 seconds
Explanation:
Applying,
T = 2usin∅/g................ Equation 1
Where, T = time of flight, u = initial velocity, ∅ = angle of projectile to the horizontal, g = acceleration due to gravity
From the question,
Given: u = 15 m/s, ∅ = 30°
Constant: g = 9.8 m/s²
Substitute these values in equation 1
T = 2(15)(sin30°)/9.8
T = 15/9.8
T = 1.53 seconds
Hence the time rate of flight is 1.53 seconds
Answer:
the answer is 3600 kilometers
