Speed = Distance/Time. Let speed be V, distance D and time T
a)Given: D₁ =45km, V₁=x km/h
V₁=D₁/T
V₁=45/T OR T= 45/V₁
b) Given : D₂ =48 km and V₂ = V₁ + 4 km/h
V₂ = 48/T, but V₂ = V₁+4, then:
V₁+4 = 48/T OR T=48/(V₁+4). Since Time is same, then we can write:
45/V₁ =48/(V₁+4), solve for V₁:
45V₁ + 180 = 48V₁
3V₁ =180 and V₁ = x = 60 km/h
Answer:
The initial speed of the car was 80 ft/s.
Step-by-step explanation:
The deceleration is the rate at which the car speed decreases. In this case the speed of the car goes all the way down to 0 ft/s and in order to do that it travelled 50 ft. So we will call the initial speed at which the car started to brake "v_0" and use Torricelli's equation to find it. The equation is given by:
v^2 = (v_0)^2 + 2*a*S
Where v is the final speed, v_0 is the initial speed, a is the rate of acceleration and S is the space travelled. Using the values that the problem gave to us we have:
0^2 = (v_0)^2 - 2*64*50
0 = (v_0)^2 - 6400
(v_0)^2 = 6400
v_0 = sqrt(6400) = 80 ft/s
Notice that in this case "a" was negative, since the car was decelerating instead of accelerating.
The initial speed of the car was 80 ft/s.
10x-6y
You have to add the like terms.
The answer to this is 1.3
