The answer to question one is A.
The answer to question two is A.
The answer to question three is D.
Answer:
d = 4 d₀o
Explanation:
We can solve this exercise using the relationship between work and the variation of kinetic energy
W = ΔK
In that case as the car stops v_f = 0
the work is
W = -fr d
we substitute
- fr d₀ = 0 - ½ m v₀²
d₀ = ½ m v₀² / fr
now they indicate that the vehicle is coming at twice the speed
v = 2 v₀
using the same expressions we find
d = ½ m (2v₀)² / fr
d = 4 (½ m v₀² / fr)
d = 4 d₀o
Check the 1st 2nd 3rd and 4th boxes
Answer:
(a) 3.807 s
(b) 145.581 m
Explanation:
Let Δt = t2 - t1 be the time it takes from the moment when the motorcycle starts to accelerate until it catches up with the car. We know that before the acceleration, both vehicles are travelling at a constant speed. So they would maintain a distance of 58 m prior to the acceleration.
The distance traveled by car after Δt (seconds) at 
speed is
The distance traveled by the motorcycle after Δt (seconds) at 
speed and acceleration of a = 8 m/s2 is
We know that the motorcycle catches up to the car after Δt, so it must have covered the distance that the car travels, plus their initial distance:
(b)
The acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.
