How much farther will a car traveling at 100 km/s skid than the same car traveling at 50 km/s?
1 answer:
Answer:
option A
Explanation:
given,
speed of car 1 = 100 Km/s
speed of car 2 = 50 Km/s
final speed = 0 Km/s
now,
skid mark when the car is moving at 100 Km/s
using equation of motion
v² = u² + 2 a s₁
0² = 100² - 2 a s₁
100² = 2 a s₁
..........(1)
now,
skid mark when the car is moving at 50 Km/s
using equation of motion
v² = u² + 2 a s₂
0² = 50² - 2 a s₂
50² = 2 a s₂
..........(2)
dividing equation (1)/ (2)
s₁ = 4 s₂
Hence, the skid mark for 100 Km/s is four time as far as 50 Km/s
The correct answer is option A
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Answer:
a) x = v₀² sin 2θ / g
b) t_total = 2 v₀ sin θ / g
c) x = 16.7 m
Explanation:
This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity
sin θ = / vo
cos θ = v₀ₓ / vo
v_{oy} = v_{o} sin θ
v₀ₓ = v₀ cos θ
v_{oy} = 13.5 sin 32 = 7.15 m / s
v₀ₓ = 13.5 cos 32 = 11.45 m / s
a) In the x axis there is no acceleration so the velocity is constant
v₀ₓ = x / t
x = v₀ₓ t
the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
t = v_{o} sin θ / g
we substitute
x = v₀ cos θ (2 v_{o} sin θ / g)
x = v₀² /g 2 cos θ sin θ
x = v₀² sin 2θ / g
at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,
b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time
at the highest point the vertical speed is zero
v_{y} = v_{oy} - gt
v_{y} = 0
t = v_{oy} / g
t = v₀ sin θ / g
as the time to get on and off is the same the total time or flight time is
t_total = 2 t
t_total = 2 v₀ sin θ / g
c) we calculate
x = 13.5 2 sin (2 32) / 9.8
x = 16.7 m
Answer:
- The procedure is: solve the quadratic equation for
.
Explanation:
This question assumes uniformly accelerated motion, for which the distance d a particle travels in time t is given by the general equation:
That is a quadratic equation, where the independent variable is the time .
Thus, the procedure that will find the time t at which the distance value is known to be D is to solve the quadratic equation for .
To solve it you start by changing the equation to the general form of the quadratic equations, rearranging the terms:
Some times that equation may be solved by factoring, and always it can be solved by using the quadratic formula:
Where:
That may have two solutions. Some times one of the solution makes no physical sense (for example time cannot be negative) but others the two solutions are valid.