Physics, 23.04.2020 01:27 MysteryDove12

A motorcycle daredevil plans to ride up a 2.0-m-high, 20° ramp, sail across a 10-m-wide pool filled with hungry crocodiles, and land at ground level on the other side. He has done this stunt many times and approaches it with confidence. Unfortunately, the motor-cycle engine dies just as he starts up the ramp. He is going 11 m/s at that instant, and the rolling friction of his rubber tires (coefficient 0.02) is not negligible. Does he survive, or does he become croco-dile food? Justify your answer by calculating the distance he travels through the air after leaving the end of the ramp

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Answer from: josebienka

He becomes a croco-dile food

Explanation:

From the question we are told that

The height is h = 2.0 m

The angle is $\theta = 20^o$

The distance is w = 10m

The speed is u = 11 m/s

The coefficient of static friction is $\mu = 0.02$

At equilibrium the forces acting on the motorcycle are mathematically represented as

$ma = mgsin \theta + F_f$

where F_f is the frictional force mathematically represented as

$F_f =\mu F_x =\mu mgcos \theta$

where F_x is the horizontal component of the force

substituting into the equation

$ma = mgsin \theta + \mu mg cos \theta$

$ma =mg (sin \theta + \mu cos \theta )$

making a the subject of the formula

$a = g(sin \theta = \mu cos \theta )$

substituting values

a = 9.8 (sin(20) + (0.02 ) cos (20 ))

= 3.54 m/s^2

Applying " SOHCAHTOA" rule we mathematically evaluate that length of the ramp as

$sin \theta = \frac{h}{l}$

making the subject

$l = \frac{h}{sin \theta }$

substituting values

$l = \frac{2}{sin (20)}$

l = 5.85m

Apply Newton equation of motion we can mathematically evaluate the final velocity at the end of the ramp as

v^2 =u^2 + 2 (-a)l

The negative a means it is moving against gravity

substituting values

v^2 = (11)^2 - 2(3.54) (5.85)

$v= \sqrt{79.582}$

= 8.92m/s

The initial velocity at the beginning of the pool (end of ramp) is composed of two component which is

Initial velocity along the x-axis which is mathematically evaluated as

v_x = vcos 20^o

substituting values

v_x = 8.92 * cos (20)

= 8.38 m/s

Initial velocity along the y-axis which is mathematically evaluated as

$v_y = vsin\theta$

substituting values

v_y = 8.90 sin (20)

= 3.05 m/s

Now the motion through the pool in the vertical direction can mathematically modeled as

$y = y_o + u_yt + \frac{1}{2} a_y t^2$

where y_o is the initial height,

u_y is the initial velocity in the y-axis

a_y is the initial acceleration in the y axis with a constant value of ( g = 9.8 m/s^2 )

at the y= 0 which is when the height above ground is zero

Substituting values

0 = 2 + (3.05)t - 0.5 (9.8)t^2

The negative sign is because the acceleration is moving against the motion

-(4.9)t^2 + (2.79)t + 2m = 0

Solving using quadratic formula

$\frac{-b \pm \sqrt{b^2 -4ac} }{2a}$

substituting values

$\frac{-3.05 \pm \sqrt{(3.05)^2 - 4(-4.9) * 2} }{2 *( -4.9)}$

$t = \frac{-3.05 + 6.9}{-9.8} \ or t = \frac{-3.05 - 6.9}{-9.8}$

$t = -0.39s \ or \ t = 1.02s$

since in this case time cannot be negative

t = 1.02s

At this time the position the motorcycle along the x-axis is mathematically evaluated as

x = u_x t

x =8.38 *1.02

x =8.54m

So from this value we can see that the motorcycle would not cross the pool as the position is less that the length of the pool

Answer from: Quest

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Answer from: Quest

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