Displacement is simply the change in position, or the difference in the final and initial positions:
Then
(a) ∆<em>d</em> = 5 m - 0 m = 5 m
(b) ∆<em>d</em> = 1 m - (-2 m) = 1 m + 2 m = 3 m
(c) ∆<em>d</em> = 2 m - (-2 m) = 2 m + 2 m = 4 m
(d) ∆<em>d</em> = 6 m - 2 m = 4 m
The velocity of projectile has 2 components, horizontal component vcosθ and vertical component vsinθ, where v is the velocity of projection and θ is the angle between +ve X-axis and projectile motion.
In case 1, θ = 90⁰
So horizontal component is vcos90 = 0
Vertical component at maximum height = 0
So velocity at maximum height = 0 m/s
In case 2, θ = 45⁰
So horizontal component is 141cos45 = 100m/s
Vertical component at maximum height = 0
So velocity at maximum height = 100 m/s
Answer:
a = 0 m/s²
Explanation:
given,
car moving at steady velocity = 100 Km/h
1 km/h = 0.278 m/s
100 Km/h = 27.8 m/s
time of acceleration = 100 s
acceleration is equal to change in velocity per unit time.
change in velocity of the car is 27.8 - 27.8 = 0
a = 0 m/s²
If the car is moving with steady velocity then acceleration of the car is zero.
Hence, the acceleration of the car is equal to a = 0 m/s²
Hey there!
When light changes speed, it REFRACTS.
Your answer is going to be option C.
Hope this helps you.
Have a great day!
Answer:
At a deceleration of 60g, or 60 times the acceleration due to gravity a person will travel a distance of 0.38 m before coing to a complete stop
Explanation:
The maximum acceleration of the airbag = 60 g, and the duration of the acceleration = 36 ms or 36/1000 s or 0.036 s
To find out how far (in meters) does a person travel in coming to a complete stop in 36 ms at a constant acceleration of 60g
we write out the equation of motion thus.
S = ut + 0.5at²
wgere
S = distance to come to complete stop
u = final velocoty = 0 m/s
a = acceleration = 60g = 60 × 9.81
t = time = 36 ms
as can be seen, the above equation calls up the given variable as a function of the required variable thus
S = 0×0.036 + 0.5×60×9.81×0.036² = 0.38 m
At 60g, a person will travel a distance of 0.38 m before coing to a complete stop
