Answer:
D = 271.54 m
Explanation:
given,
1. car accelerates at 4.6 m/s² for 6.2 s
2. constant speed for 2.1 s
3. slows down at 3.3 m/s²
distance travel for case 1
using equation of motion
d₁ = 88.41 m
case 2
constant speed for 2.1 s now, we have to find velocity
v = u + at
v = 0 + 4.6 x 6.2
v = 28.52 m/s
distance travel in case 2
d₂ = v x t
d₂ = 28.52 x 2.1 = 59.89 m
for case 3
distance travel by the car
v² = u² + 2 a s
final velocity if the car is zero
0² = 28.52² + 2 x (-3.3) x d₃
6.6 d₃ = 813.39
d₃ = 123.24 m
total distance travel by the car
D = d₁ + d₂ + d₃
D = 88.41 + 59.89 + 123.24
D = 271.54 m
The watt<span> (symbol: W) is a unit of power i hope this helps you</span>
nuclear ... truman, hiroshima, nagasaki
Answer:
-414.96 N
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
The force the ground exerts on the parachutist is -414.96 N
If the distance is shorter than 0.75 m then the acceleration will increase causing the force to increase
Answer:
6.429 m/s^2.
Explanation:
Using equations of motion,
i. vf = vi + at
ii. vf^2 = vi^2 + 2a*S
iii. S = vi*t + 1/2 * (a*t^2)
Where,
vf = final velocity of the motion
vi = initial velocity of the motion
S = distance travelled
t = time taken to complete the motion
a = acceleration due to gravity
Given:
vi = 0m/s
vf = 45 m/s
t = 7 s
a = ?
Using the i. equation of motion,
vf = vi + at
45 = 0 + a*7
a = 45/7
= 6.429 m/s^2
