Unit B: Changes in Motion

The reaction distance is the distance a vehicle travels while the driver is reacting to some condition or hazard. Since the vehicle is travelling with uniform motion, the reaction distance is equal to the speed of the vehicle multiplied by the reaction time of the driver.

Braking distance is the distance the vehicle travels from the instant the driver’s foot touches the brake pedal to the time the vehicle stops. Because the vehicle is decelerating, the calculation of the braking distance is a two-step process:

step 1: Determine the time it takes for the vehicle to stop.

step 2: Calculate the distance using one of the accelerated motion equations.

Stopping distance is the sum of the reaction distance and the braking distance.

An applied force is an external force applied to an object. If you were pulling a toboggan across the snow, the force you apply through the rope is the applied force on the toboggan.

The force of friction is a contact force between two surfaces that acts to oppose the motion of one surface past the other. To return to the example of pulling a toboggan across the snow, the force of friction is the force between the bottom of the toboggan and the snow that pushes in the opposite direction of the toboggan’s motion.

The net force is the vector sum of all the forces acting on the toboggan. When a toboggan is being pulled across the snow, the net force is the force that remains once the force of friction has been subtracted from the applied force.

Newton’s first law states that, in the absence of a net force, an object in motion will tend to maintain its velocity and an object at rest will tend to remain at rest. Newton’s first law is sometimes called the law of inertia. Newton’s first law explains why an unrestrained passenger in a vehicle will continue to move in the original direction even though the vehicle may have been stopped abruptly.

Newton’s second law states that an object will tend to accelerate in the direction of an unbalanced force such that Newton’s second law explains why a force is required to accelerate a vehicle.

Starting at the corner of Bay and Wellington, the distance to the Capital Information Centre is 725 m. The position of the Capital Information Centre is 725 m, east. The position is the better description because it also includes the direction of how to get there.

The velocity is 3.63 km/h [E].

The resultant displacement is 395 m [E].

The average speed is 3.72 km/h.

Determine the average velocity.

The average velocity is 1.39 km/h [E].

The average velocity has a smaller magnitude because it is based on a smaller value for displacement. This is due to the fact that one displacement is east (positive direction) while the other is west (negative direction).

Note: Since this graph describes uniform motion, an accelerated motion equation is not required to determine displacement.

The motion of the car in the section labelled “reaction distance” is uniform motion. By Newton’s first law, the car is able to maintain its velocity in this section because the forces on the car are balanced. This motion will not last long because as soon as the driver pushes on the brake pedal, the forces on the car will become unbalanced.

The motion of the car in the section labelled “braking distance” is negatively accelerated, or decelerated motion. The car had an initial velocity of 96.0 km/h, east, and a final velocity of zero; therefore, the car was decelerating. Newton’s laws explain this further. The negative acceleration means that the acceleration vector was pointing in the negative (or westward) direction due to a net force, which was also acting in the negative (or westward) direction. This net force was negative because the forces that caused the braking must push opposite to the original direction of the car, which was positive.

The stopping distance is 156 m.

If this emergency stopping manoeuvre had happened at night, the driver would not have been able to stop in time. As the previous answer indicates, the driver needed 156 m to stop the vehicle, so even with the added illumination of the HID headlights, the driver would not have been able to stop.

The speed limit is not the speed that you are supposed to travel. The speed limit is the maximum speed for a vehicle under optimum conditions. The fact that this happened at night (reducing visibility) and the fact that the road was slippery (reducing the ability to decelerate) meant that the driver should have been travelling at a velocity less than 96 km/h.

The final momentum is 0.

The final velocity of the truck is 13 m/s [N].

The final velocity of the combined mass is 8.0 m/s [N].

No, a primary collision will not cause injury to a passenger in a vehicle. The primary collision is between the vehicle and the obstacle. The secondary and tertiary collisions will cause injury to the occupant.

Since and since the is the same in both cases, F must decrease as increases. Letting your hand “give” increases the time, thus making catching the ball less painful because of the decreased force.

The force exerted by his arm on the ball over 1.2 s is 7.0 N [backward].

Rigid Arms

The computer image on the far right best shows a collision that reduces the forces due to the longer stopping time. The computer image in the centre shows the larger forces acting (ten times larger) because the stopping time is so small (ten times smaller).

Car

The truck travelled 23.7 m while braking.

The car travelled 34.2 m while braking.

The truck’s stopping distance is 48.7 m.

The car’s stopping distance is 64.2 m.

For Ramp A, the skateboard is rolling downhill, speeding up as it goes.

For Ramp B, the skateboard is rolling uphill, slowing down as it goes. Because the magnitude of the acceleration is slightly greater for Ramp B, this ramp must be a little steeper.

The type of motion depicted by the graph for the truck is uniform motion because the graph is a straight line.

m1 = 1200 kg
m2 = 900 kg

The impulse on the truck is 6.27 × 10³ N•s [W].

Yes, the law of conservation of momentum states that the magnitudes of the change in momentums must be the same. However, one object is losing momentum and the other object is gaining momentum.

The impulse of the truck on the car is 8.88 × 10³ N•s [E].

The impulse of the car on the truck is 8.88 × 10³ N•s [W].

To do this work, 2.05 × 10³ J of energy is required.

No, forensic engineers do not determine who was at fault in a collision. Forensic engineers collect and analyze data to try to determine exactly what happened.