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Kay [80]
2 years ago
9

A boy runs 40m towards east. he then walks 30m towards north. The boy again runs 50m towards north making an angle of 30° with e

ast. What is his displacement?​
Physics
1 answer:
Katena32 [7]2 years ago
5 0

Answer: The displacement is 99.82 meters.

Explanation:

To use a standard notation, we can define:

North as the positive y-axis.

East as the positive x-axis.

If the initial position of the boy is (0, 0)

Then:

"runs 40m towars the east"

The new position is (0, 0) + (40m, 0) = (40m, 0)

"he then walks 30m towards North"

The new position is: (40m, 0) + (0, 30m) = (40m, 30m)

"The boy again runs 50m towards north making an angle of 30° with East."

By making a triangle rectangle, we can find the x and y components for this motion.

Y-component = 50m*cos(30°) = 43.3m

X-component = 50m*sin(30°) = 25m

Then the new position will be:

(40m, 30m) + (43.3m, 25m) = (83.3m, 55m)

Now we want to find the total displacement, this will be equal to the distance between the final position and the initial position.

The distance between two points (ax, ay) and (bx, by) is:

D = √( (ax - bx)^2 + (ay - by)^2)

In this case we want to find the distance between the points  (83.3m, 55m) and (0, 0), this will be:

D = √( (83.3m - 0)^2 + (55m - 0)^2) = 99.82 m

The displacement is 99.82 meters.

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The net horizontal force on a car is 981 N. The car has a mass of 1550 kg and the force is applied when the car has a speed of 2
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Distance, d = 778.05 m                          

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Given that,

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We need to find the distance covered by car if the force continues to be applied to the car. Firstly, lets find the acceleration of the car:

F=ma\\\\a=\dfrac{F}{m}\\\\a=\dfrac{981}{1550}\\\\a=0.632\ m/s^2

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d=ut+\dfrac{1}{2}at^2\\\\d=11.17\times 35+\dfrac{1}{2}\times 0.632\times (35)^2\\\\d=778.05\ m

So, the car will cover a distance of 778.05 meters.

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