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2 years ago

At noon, Bradley began steadily increasing the

Mathematics

2 answers:

barxatty [35]2 years ago

8 0

Answer:

25.5 mph

Step-by-step explanation:

So Bradley's speed can be modeled by the equation y=2x+40 where y=speed, x=time in hours after noon, and b=initial speed

So 12:15 is 15 minutes after noon, which is also 0.25 or 1/4 of an hour after noon. This is the x-value. Plug this into the equation to get his speed at 12:15

y=2(0.25)+40

y=0.5+40

y=40.5

So his speed was 40.5 at the time and since he was going 15 miles over the speed limit, the speed limit is 15 less than his speed

40.5 - 15 = 25.5

Colt1911 [192]2 years ago

5 0

Answer:

25.5 mph

Step-by-step explanation:

Given information:

Bradley increases the speed of his car by 2 miles per hour.
12.00 pm: Bradley's speed = 40 mph
12.15 pm: Bradley is driving at 15 miles over the speed limit.

Calculate Bradley's speed at 12.15 pm:

60 minutes = 1 hour

⇒ 15 minutes = 1/4 hour

⇒ speed at 12.15 pm = speed at 12.00pm + 1/4 of 2 mph

= 40 mph + 0.5 mph

= 40.5 mph

If he was 15 miles over the speed limit at 12.15 pm then:

⇒ speed limit = speed at 12.15pm - miles over the limit

= 40.5 mph - 15

= 25.5 mph

Therefore, the speed limit was 25.5 mph.

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3 years ago

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3 years ago

Read 2 more answers

Doug hits a baseball straight towards a 15 ft high fence that is 400 ft from home plate. The ball is hit 2.5ft above the ground

Scorpion4ik [409]

Answer:

$125.4\ \text{m/s}$

Step-by-step explanation:

u = Initial velocity of baseball

$\theta$ = Angle of hit = $30^{\circ}$

x = Displacement in x direction = 400 ft

y = Displacement in y direction = 15 ft

$y_0$ = Height of hit = 2.5 ft

$a_y$ = g = Acceleration due to gravity = $32.2\ \text{ft/s}^2$

t = Time taken

Displacement in x direction

$x=u_xt\\\Rightarrow x=u\cos\theta t\\\Rightarrow t=\dfrac{x}{u\cos\theta}\\\Rightarrow t=\dfrac{400}{u\cos30^{\circ}}\\\Rightarrow t=\dfrac{400}{u\dfrac{\sqrt{3}}{2}}\\\Rightarrow t=\dfrac{800}{u\sqrt{3}}$

Displacement in y direction

$y=y_0+u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow y=y_0+u\sin\theta t+\dfrac{1}{2}a_yt^2\\\Rightarrow 15=2.5+u\sin30^{\circ}(\dfrac{800}{u\sqrt{3}})+\dfrac{1}{2}\times -32.2\times (\dfrac{800}{u\sqrt{3}})^2\\\Rightarrow \dfrac{400}{\sqrt{3}}-\dfrac{10304000}{3u^2}-12.5=0\\\Rightarrow 218.44=\dfrac{10304000}{3u^2}\\\Rightarrow u=\sqrt{\dfrac{10304000}{3\times218.44}}\\\Rightarrow u=125.4\ \text{m/s}$

The minimum initial velocity needed for the ball to clear the fence is $125.4\ \text{m/s}$

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