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blagie [28]
2 years ago
13

A boat is stationary at 12 meters away from a dock. The boat then begins to move toward the dock with

Physics
1 answer:
just olya [345]2 years ago
8 0

Answer: 2.19 seconds

Explanation:

<u>Given:</u>

Initial speed, u = 0 m/s

Acceleration of the boat, a = 5 m/s^2

Distance between the boat and dock, d = 12 m

Using the third kinematics equation to solve for time:

d = u*t + (1/2)*a*t^2

12 = 0*t + (1/2)*5*t^2

t = sqrt (12*2/5)

t = 2.19 seconds

Therefore, it will take the boat approximately 2.19 seconds to reach the dock

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Two cars are traveling along a straight line in the same direction, the lead car at 25 m/s and the other car at 35 m/s. At the m
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Answer:

a. t_1=12.5\ s

b. a_2=-13.61\ m.s^{-2}  must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c. t_2=2.5714\ s is the time taken to stop after braking

Explanation:

Given:

  • speed of leading car, u_1=25\ m.s^{-1}
  • speed of lagging car, u_{2}=35\ m.s^{-1}
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a.

Time taken by the car to stop:

v_1=u_1+a_1.t_1

where:

v_1=0 , final velocity after braking

t_1= time taken

0=25-2\times t_1

t_1=12.5\ s

b.

using the eq. of motion for the given condition:

v_2^2=u_2^2+2.a_2.\Delta s

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v_2= final velocity of the chasing car after braking = 0

a_2= acceleration of the chasing car after braking

0^2=35^2+2\times a_2\times 45

a_2=-13.61\ m.s^{-2} must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c.

time taken by the chasing car to stop:

v_2=u_2+a_2.t_2

0=35-13.61\times t_2

t_2=2.5714\ s  is the time taken to stop after braking

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