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STatiana [176]
3 years ago
7

At the instant the traffic light turns green, a car starts with a constant acceleration of 8.00 ft/s^2. At the same instant a tr

uck, traveling with a constant speed of 10.0 ft/s, overtakes and passes the car. How far from the starting point (in feet) will the car overtake the truck?
Physics
1 answer:
kompoz [17]3 years ago
7 0

Answer:

they meet at distance 25 feet

Explanation:

given data

acceleration of car  = 8 ft/s²

truck speed = 10 ft/s

car initial speed u = 0

truck acceleration = 0

to find out

How far from the starting point will car overtake the truck

solution

we apply here equation of motion

s = ut + 0.5 ×a×t²   .............1

here s is distance and a is acceleration and t is time u is initial speed

so truck distance

s = 10t + 0.5 ×0×t²

s = 10 t   ...............2

and car distance

s = 0+ 0.5 ×8×t²  

s = 4×t²     ..........................3

so from equation 2 and 3

10 t = 4×t²

t = 2.5 s

so both meet at distance

s = 10 (t)

s = 10 ( 2.5 ) = 25 ft

so they meet at distance 25 feet

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GREYUIT [131]

Answer:

303.29N and 1.44m/s^2

Explanation:

Make sure to label each vector with none, mg, fk, a, FN or T

Given

Mass m = 68.0 kg

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g = 9.8m/s^2

Coefficient of static friction μs = 0.50

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Solution

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N = mg - Fsinθ

Horizontally

Fs = F cos θ

μsN = Fcos θ

μs( mg- Fsinθ) = Fcos θ

μsmg - μsFsinθ = Fcos θ

μsmg = Fcos θ + μsFsinθ

F = μsmg/ cos θ + μs sinθ

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F = 332.2/0.9659+0.5×0.2588

F =332.2/1.0953

F = 303.29N

Fnet = F - Fk

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a = 303.29 - 0.35(68.0 * 9.8- 303.29*sin15)/68.0

303.29-0.35( 666.4 - 303.29*0.2588)/68.0

303.29-0.35(666.4-78.491)/68.0

303.29-0.35(587.90)/68.0

(303.29-205.45)/68.0

97.83/68.0

a = 1.438m/s^2

a = 1.44m/s^2

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3 years ago
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Answer:
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Explanation
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Answer:

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