Find the time required for the boat to slow to 45 km/hr

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The discussion focuses on calculating the time required for a 1000 kg boat to decelerate from 90 km/h to 45 km/h after its engine is turned off. The frictional force acting on the boat is defined as 70v, where v is the speed in meters per second. Using Newton's law, the time required for the boat to slow down is determined to be approximately 14.29 seconds, derived from the formula t = (v - u) / a, where the acceleration is calculated as 0.875 m/s².

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hey

A 1000 kg boat is traveling at 90km/h when its engine is shut off. The magnitude of the frictional force between boat and water is proportional to the speed v of the boat: force of kinetic fricition=70v, where v is in meters per second and the force is in Newtons. Find the time required for the boat to slow to 45 km/hr
 
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king_naeem said:
hey

A 1000 kg boat is traveling at 90km/h when its engine is shut off. The magnitude of the frictional force between boat and water is proportional to the speed v of the boat: force of kinetic fricition=70v, where v is in meters per second and the force is in Newtons. Find the time required for the boat to slow to 45 km/hr

I'm a theorist,so i'll let u plug the numbers:
The Newton's law states:m\frac{dv}{dt}=-\alpha v,where alpha is that 70 in your problem:
Integrate the ODE wrt to proper limits:
Final result:t=(\frac{1000}{70}\ln 2) s

Good luck!
 
.

To find the time required for the boat to slow to 45 km/hr, we can use the formula for acceleration: a = F/m, where a is the acceleration, F is the net force, and m is the mass of the boat. In this case, the net force is the frictional force, which is given by 70v. We can also convert the speed of the boat from km/hr to m/s by multiplying it by 1000/3600 (since 1 km/hr = 1000/3600 m/s). So, the speed of the boat in m/s is 45 * 1000/3600 = 12.5 m/s.

Now, we can plug in the values into the formula for acceleration:

a = 70 * 12.5 / 1000 = 0.875 m/s^2

Next, we can use the formula for uniform acceleration to find the time required for the boat to slow down from 12.5 m/s to 0 m/s:

v = u + at

Where v is the final velocity (0 m/s), u is the initial velocity (12.5 m/s), a is the acceleration (0.875 m/s^2) and t is the time we are looking for. Rearranging the equation to solve for t, we get:

t = (v-u)/a = (0 - 12.5)/0.875 = -14.29 seconds

However, we cannot have a negative time, so we can conclude that it will take approximately 14.29 seconds for the boat to slow down from 90 km/hr to 45 km/hr.
 

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