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leva [86]
3 years ago
9

A bug is moving along a meter stick in the negative direction at a constant speed of 0.85cm/s. After 42 seconds have passed the

bug is located at 27.0 cm. What was the bug’s starting position.
Physics
1 answer:
Agata [3.3K]3 years ago
3 0

Given that,

bug speed, v= 0.85 m/s

time, t =42 s

Final position of bug on meter stick was 27 cm

Starting position of bug on meter stick = ?

Since we know that,

s = vt

s= 0.85*42 = 35.7 cm

this is the distance covered by bug in the given time and velocity.

since the bug is moving in negative direction, starting point will be:

27.0 cm+ 35.7 cm = 62.7 cm

The bugs starting position on meter stick was 62.7 cm.

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(a)The acceleration of the 1,400-kg boat will be 0.425 m/sec²

(b) If it starts from rest, the distance through which the boat moves in 20.0 s will be 85 m.

(c)Velocity at the end of that time will be 8.5 m/sec.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

Given data;

Forwad force acting on the boat,v₁ = 1,800-N

Resistive force acting on the boat,v₂ = 1,200-N

The acceleration of the boat,a

Mass of boat,m = 1,400-kg

Initial velocity of boat,u= 0 m/sec

Distance travelled by boat,S = ?

Time for the boat travels,t = 20.0 s

Final velocity,V = ? m/sec

The net force on the boat;

F = F₁ - F₂

F = 1800 N - 1200 N

F = 600 N

From the defination of force;

F= ma

a = F / m

a = 600 N / 1400 kg

a = 0.425 m/sec²

b)

The distance through which the boat moves is 20.0 s;

\rm x_f = x_ 0 + v_0 t + \frac{1}{2} at^2 \\\\ x_f  = \frac{1}{2}at^2 \\\\ x_f = 0+0 + \frac{1}{2} at^2 \\\\ x_f = \frac{1}{2}\times 0.425 \times (20 )^ 2 \\\\ x_f  = 85 \ m

c)

The velocity at the end of that time is found as;

\rm  v_f = v_ 0 + at \\\\ v_f =- 0+ at \\\\ v_f = 8.5 m/sec

Hence the acceleration of the boat, the distance through which the boat moves in 20.0 s, and velocity at the end of that time will be 0.425 m/sec²,85 m, and 8.5 m/sec.

To learn more about the velocity, refer to the link: brainly.com/question/862972

#SPJ1

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