a.)We first use kinematics to find the car's acceleration, remembering to put everything into SI units...

b.)From Newton's Second Law, the net force = F = Ma, where M is the mass of the car (500 kg) and a is the car's acceleration (-5.6 m/s^{2}). So

F = (500 kg)(-5.5 m/s^{2}) = -2.8 x 10^{3}kg.m/s^{2}= -2.8 x 10^{3}Nc.)The net force is applied at the only point of contact the car has -- the ground! The ground pushes on the car opposite to the car's motion, causing it to stop. You can tell this, because a car's tires get hot from the friction forces between them and the road.

d.)Constant acceleration means a linear variation of velocity with time

From the graph, it's clear that the average velocity over the time while the car is stopping is 40 km/hr. This means that the distance traveled by the car while it is stopping is

distance = (average velocity) (stopping time)

distance = (40 km/hr) (4 s) = (11 m/s)(4 s) = 44 m.

a.)The ball has a constant acceleration of 9.8 m/s^{2}downward (the acceleration due to gravity). This means a linear velocity versus time graph (for a similar picture, see the previous problem). The ball is "dropped" from the top of the cliff, implying that its initial velocity is zero. So, after 4 seconds...velocity after 4 seconds = (9.8 m/s^{2})(4 s) = 39 m/s

b.)Again, it might help to look at a velocity versus time graph for an object moving with constant acceleration (see the solution to S8). The point is, that over the 4 s time interval, theaverage velocityof the ball is half of itsinstantaneousvelocity at 4 s, namely 19.6 m/s. Armed with that, we can get the distance fallen

distance fallen = (avg velocity) (time) = (19.6 m/s) (4 s) = 78 m.c.)From Newton's Second law, the net force on the ball = F = Ma, where M is the mass of the ball (1.0 kg). This means that the ball has a force on it = (1.0 kg)(9.8 m/s^{2}) = 9.8 N. This force is the force due to gravity, or the ball's weight.

S9Click here to read the questiona.)The first thing we need to do is to understand the motion of the ball. There is amaximum heightthat the ball reaches where the velocity is zero. The other essential aspect to the motion is that the ball has exactly the same speed on landing as it did at it's release point, when it was thrown upward. The only difference is that the velocity is -15 m/s upon landing compared to +15 m/s when it is initially thrown. Armed with knowledge that the upward motion and downward motion aresymmetricit should be clear that the totaltime of flightis simply twice the time (T) for the ball to reach its maximum height!

For constant acceleration due to gravity, a = -9.8 m/s^{2}, the time to reach the maximum height is the time it takes for the ball to instantaneously come to rest or,delta_v = vThis means that T = delta_v / a = (-15 m/s) / (-9.8 m/s_{max height}- v_{initial}= 0 - 15 m/s.^{2}) = 1.5 s

The total time of flight is 2T = 3.0 s.b.)We've already answered that from the symmetry of the motion: at landing, the ball's velocity is -15 m/s. Perhaps a graph of velocity vs. time will help?

c.)The ball's acceleration is -9.8 m/s^{2}. From Newton's second law, the net force on the ball is F = Ma = (1.0 kg)(-9.8 m/s^{2}) = -9.8 N. The minus sign simply means that the force points in the same direction as the acceleration, downward! The force is the weight of the ball.d.)The force on the ball is the same at all points along its path.Rossing 1.6An object weighing one pound (English units) has a mass of 0.455 kg. Express its weight in newtons and thereby express a conversion factor for pounds to newtons.An object with a mass (M) of 0.455 kg has a weight,W = Mg = (0.455 kg)(9.80 m/sSo, our units conversion factor is^{2}) = 4.46 N.1 pound = 4.46 NReturn to P105 Course Schedule

Last updated:

10 Sep 1999Comments: bland@indiana.edu