Example Problems for Kinematics. II

  1. The nucleus of a helium atom (an alpha particle) travels along the inside of a straight hollow tube 2.0 m long which forms part of a particle accelerator.
         (a). Assuming uniform acceleration, for how long is the particle in the tube if it enters at a speed of 1.0 x 104 m/sec and leaves at 5.0 x 106 m/sec?
         (b). What is its acceleration during this interval?

  2. The speed of an automobile traveling due east is uniformly reduced from 75.0 km/hr (=46.6 mi/hr) to 45.0 km/hr (=28 mi/hr) in a distance of 88.5 m (=290 ft).
         (a). What is the magnitude and direction of the constant acceleration?
         (b). How much time elapses during this deceleration?
         (c). If we assume that the car continues to decelerate at the rate calculated in (a), how much time would elapse in bringing it to rest from 75.0 km/hr?

  3. Suppose I throw an eraser upward at 5.40 m/sec. How long will it take to return to my hand? How high will it go? What will its acceleration be at the top of its path? Neglect air resistance.

  4. Suppose I throw an eraser at an angle of 60 degrees above the horizontal at a speed such that the upward component of the velocity is 5.40 m/s. How far will the eraser travel before returning to the height at which it was released? How high will it go?

 

Solution to Problem 3
Diagram of
y vs. t
(a parabola) 		v0 = 5.40 m/s
a = -g = 9.8 m/sec2
t2 = ?
ymax = ?

The situation is one of motion in one dimension with a constant acceleration of  -g [g downward], so

y = y0 + v0 t - (1/2) g t2
v = v0 - g t

Both y0 and y(t2) = 0, and we can find the flight time t2 from

0 = v0 t2 - (1/2) g t2 2
t2 = 2 v0 / g = 1.1 sec

Finding the maximum height is a separate problem. The maximum is given by calculus as

dy/dt = 0

Rewriting this equation in terms of  v = dy/dt gives

v = 0

and the working equations become

0 = v0 - g t1
ymax = 0 + v0 t1 - (1/2) g t12

The first of these equations may be solved to give

t1 = v0 / g = (1/2) t2

Hence exactly half of the flight time is used going up, and of course the remaining half is used to come back down. Substituting the known value of t2 into the equation for y gives

ymax = v02 / g - (1/2)g (v0 / g )2
ymax = v02 / (2 g) = 1.49 m