Question

In: Physics

Suppose a rocket-propelled motorcycle is fired from rest horizontally across a canyon 1.20 km wide. (a)...

Suppose a rocket-propelled motorcycle is fired from rest horizontally across a canyon 1.20 km wide.

(a) What minimum constant acceleration in the x-direction must be provided by the engines so the cycle crosses safely if the opposite side is 0.780 km lower than the starting point?
m/s²

(b) At what speed does the motorcycle land if it maintains this constant horizontal component of acceleration? Neglect air drag, but remember that gravity is still acting in the negative y-direction.
m/s

Solutions

Expert Solution

(a) For the motorcycle to land safely, the displacement it undergoes in the y-direction should be equal to the difference between the heights of the starting point and landing point, h.

We have, h = 0.78 km = 780 m

The acceleration in the y-direction is equal to the acceleration due to gravity (as there is no external component of acceleration provided in the y-direction)

The motorcycle starts from rest, so its initial velocity, u_y = 0.

Using the above values and the 2nd kinematic equation of motion, we can calculate the time taken for the motorcycle to descend 780 m with an acceleration of 9.8 m/s²

h = 780 = u_yt + gt² / 2 = 0 + 9.8*t²/2

t² = 780/4.9

t = 12.62 s

Now, we have the horizontal distance that the bike has to cover, s = 1.2 km = 1200 m.

The bike starts from rest (u_x = 0) and we need to know the constant acceleration a that we need to provide to the bike in order for it to safely cross the canyon. Using the same equation as we did above

s = 1200 = u_xt + at² / 2 = 0 + a*(12.62)² / 2

a = 2*1200 / 12.62² = 15.08 m/s²

Ans. We need to provide an acceleration of 15.08 m/s² in the x-direction for the motorcycle to land safely.

(b) We have the horizontal and vertical components of acceleration

a_x = a = 15.08 m/s²

a_y = g = 9.8 m/s²

The above two accelerations are constant.

We also have the time taken for the motorcycle to cross the canyon.

t = 12.62 s.

We can use these to calculate the horizontal and vertical components of the velocity of the motorcycle as it lands.

Using the 1st kinematic equation of motion.

v_x = u_x + a_xt = 0 + at = 15.08 * 12.62 = 190.31 m/s

v_y = u_y + a_yt = 0 + gt = 9.8 * 12.62 = 123.68 m/s

The speed of the motorcycle can now be calculated

v = (v_x² + v_y²)^1/2 = 226.97 m/s

Ans. The motorcycle lands with a speed of 226.97 m/s.

genius_generous answered 1 year ago

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