Answer:
Distance, d = 778.05 m
Explanation:
Given that,
Force acting on the car, F = 981 N
Mass of the car, m = 1550 kg
Initial speed of the car, v = 25 mi/h = 11.17 m/s
We need to find the distance covered by car if the force continues to be applied to the car. Firstly, lets find the acceleration of the car:

Let d is the distance covered by car. Using second equation of motion as :

So, the car will cover a distance of 778.05 meters.
Answer:
a=0 v = v₀ + a t
a=0 line is horizontal
Explanation:
1, In a graph of acceleration vs. time, we have lines, when the line is horizontal it is zero, when the line has a positive slope the increasing accelerations and when the slope is negative the decreasing acceleration
2, speed and relationship of a car is given by
v = v₀ + a t
where vo is the initial velocity, a is the acceleration and tel time
in this case I will calcograph velocity vs. time the constant acceleration is a straight line.
In general from the graph we can find the initial velocity with the cut at that x and the acceleration of the car with the slope
Answer:
17.2 seconds
Explanation:
Given:
v₀ = 0 m/s
a₁ = 10.0 m/s²
t₁ = 3.0 s
a₂ = 16 m/s²
t₂ = 5.0 s
a₃ = -12 m/s²
v₃ = 0 m/s
Find: t
First, find v₁:
v₁ = a₁t₁ + v₀
v₁ = (10.0 m/s²) (3.0 s) + (0 m/s)
v₁ = 30 m/s
Next, find v₂:
v₂ = a₂t₂ + v₁
v₂ = (16 m/s²) (5.0 s) + (30 m/s)
v₂ = 110 m/s
Finally, find t₃:
v₃ = a₃t₃ + v₂
(0 m/s) = (-12 m/s²) t₃ + (110 m/s)
t₃ = 9.2 s
The total time is:
t = t₁ + t₂ + t₃
t = 3.0 s + 5.0 s + 9.2 s
t = 17.2 s
Round as needed.
Answer: 0.8 m
Explanation:
Velocity of throw = 4m/s
Maximum Height attained(h) =?
Downward acceleration experienced = 10m/s^2
Using the relation:
v^2 = u^2 + 2aS
v = final Velocity = 0 (at maximum height)
u = Initial Velocity = 4
a = g downward acceleration = - 10
0 = 4^2 + 2(-10)(S)
0 = 16 - 20S
20S = 16
S = 16 / 20
S = 0.8m
Maximum Height attained = 0.8m