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maw [93]
4 years ago
10

Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc

e DA beyond the starting line at t=0. The starting line is at x=0. Car A travels at a constant speed vA. Car B starts at the starting line but has a better engine than Car A, and thus Car B travels at a constant speed vB, which is greater than vA
How long after Car B started the race will Car B catch up with Car A?
Physics
1 answer:
kumpel [21]4 years ago
7 0

Answer:\frac{D_A}{v_B-v_A}

Explanation:

Given

car A had a head start of D_A

and it starts at x=0 and t=0

Car B has to travel a distance of D_A and d_a

where d_a is the distance travel by car A in time t

distance travel by car A is

d_a=v_A\times t

For car B with  speed v_B

d_B=D_A+d_a

v_B\times t=D_A+v_A\times t

t=\frac{D_A}{v_B-v_A}

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Desde el balcón de un edificio se deja caer una manzana y llega a la planta baja en 5 s. ¿Desde qué piso se dejó caer, si cada p
miv72 [106K]

Answer: 42

Explanation:

I will answer this in English.

We know that the apple needs 5 seconds to reach the ground.

Each floor of the building has a height of 2.88m.

Now, when we drop something, the only force acting on the object is the gravitational one, so the acceleration of the apple is:

a = -g

for the velocity, we integrate the acceleration over time, and as the apple is dropped, we do not have any initial velocity, so we do not have a constant in the integration:

v = -g*t

for the position we integrate again, now we have an initial height H, so the position is:

p = (-g/2)*t^2 + H

now the apple hits the ground when p = 0, so we can solve this equation to find H.

i will use g = 9.8m/s^2

0 = (-4.9m/s^2)*(5s)^2 + H  

H = 122.5 m

now knowing H, we can divide it by the height of a floor in the building and get the number of the floor.

N = 122.5m/2.88m = 42.5

this means that the apple was dropped in the floor 42 (the 0.5 means that the apple was not right where the floor 42 starts, it was dropped around the middle of the floor 42)

5 0
3 years ago
With speeds up to 90 miles per hour, what is the fastest sport on ice?.
Eddi Din [679]

Answer:

Luge

Explanation:

8 0
3 years ago
Two boats start together and race across a 48-km-wide lake and back. boat a goes across at 48 km/h and returns at 48 km/h. boat
jolli1 [7]

Answer:

Time required by boat 1 for the round trip is less than that of boat 2.

Hence, boat 1 wins.

Explanation:

Case 1: Boat 1

Speed of boat = \frac{distance of river}{time}

time = \frac{distance of river}{speed of boat}

While going to another end

time = \frac{distance of river}{speed of boat}

time = \frac{48}{48}

time = 1 hour

While going back,

time = \frac{distance of river}{speed of boat}

time = \frac{48}{48}

time = 1 hour

Total time taken by boat 1 is,

Total time by boat 1 = 1 hour + 1 hour = 2 hour

Total time by boat 1 = 2 hour

Total time taken by boat 1 for the round trip is 2 hour.

Case 2: Boat 2

Speed of boat = \frac{distance of river}{time}

time = \frac{distance of river}{speed of boat}

While going to another end

time = \frac{distance of river}{speed of boat}

time = \frac{48}{24}

time = 2 hour

While going back,

time = \frac{distance of river}{speed of boat}

time = \frac{48}{72}

time = 0.66 hour

Total time taken by boat 2 is,

Total time by boat 1 = 2 hour + 0.66 hour

Total time by boat 1 = 2.66 hour

Total time taken by boat 2 for the round trip is 2.66 hour.

Time required by boat 1 for the round trip is less than that of boat 2.

Hence, boat 1 wins.

5 0
3 years ago
The graphs display velocity data Velocity is on the y-axis (m/s), while time is on the x-axis (s). Based on the graphs, which da
RUDIKE [14]

Answer:

The first graph is showing the constant acceleration (1 m/s)

Explanation:

The second graph showing the flexible velocity therefore a in the graph is different at t1, t2, t3, t4

The last graph is showing constant velocity therefore there is no acceleration (a = 0)

5 0
3 years ago
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