Question

A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to try leaping it with his car. The side the car is on is 20.8 mm above the river, whereas the opposite side is a mere 1.3 mm above the river. The river itself is a raging torrent 53.0 mm wide.
A) How fast should the car be traveling just as it leaves the cliff in order to just clear the river and land safely on the opposite side?
B) What is the speed of the car just before it lands safely on the other side?

Answer 1

Answer:

A) 26.5 m/s

B) 33.0 m/s

Explanation:

A)

• Once the car leaves the cliff, as no other influence than gravity acts on it, and since it causes the car an acceleration in the vertical direction only, in the horizontal direction, it keeps moving at the same speed until it reaches to the other side.
• So, we can apply the definition of average velocity to find this speed as follows:

       $v_{x} = \frac{\Delta x}{\Delta t} (1)$

• We know the value of Δx, which is just the wide of the river (53.0m), but we need to find also the value of Δt.
• This time is given by the vertical movement, whic.h is independent from the horizontal one, because both movements are perpendicular each other.
• Since the only influence in the vertical direction is due to gravity, the car is accelerated by gravity, with constant acceleration downward equal to g = -9.8m/s² (taking the upward direction as positive).
• Since the acceleration is constant, we can use the following kinematic equation, as follows:

       $\Delta y = y_{f} - y_{o} = v_{o} * t + \frac{1}{2} * g *t^{2} (2)$

• if we take the river level as our x-axis, this means that yf = 1.3 m and

       y₀ = 20.8 m.

• At the same time, due to in the vertical direction the car has no initial velocity, this means that  v₀ = 0.
• Replacing by the values in (2) , and solving for t:

       $t = \sqrt{\frac{2* \Delta y}{g} } = \sqrt{\frac{2*19.5m}{9.8m/s2} } = 2 s (3)$

• If we choose t₀ =0 ⇒ Δt = t = 2 s
• Replacing Δx and Δt in (1):

       $v_{x} = \frac{\Delta x}{\Delta t} = \frac{53.0m}{2s} = 26.5 m/s (4)$

B)

• When the car is just landing in the other side, the velocity of the car has two components, the horizontal one that we just found in A) and a vertical one.
• Due to the initial velocity in the vertical direction was just zero, we can find the final velocity just applying the definition of acceleration, with a =g, as follows:

      $v_{fy} = g*t = -9.8m/s2*2 s = -19.6 m/s (5)$

• Since both components are perpendicular each other, we can find the magnitude of the velocity vector (the speed) using the Pythagorean Theorem, as follows:

       $v = \sqrt{v_{x}^{2} + v_{fy}^{2} } } = \sqrt{(26.5m/s)^{2} + (-19.6m/s)^{2}} = 33.0 m/s (6)$