A car is travelling at 50km / h The driver sees a child run out into the road 5 m ahead. She applies the breaks and the car stops in 5 seconds. The driver's thinking time is 1.5 s.

a) Will the car stop in time?

b) If the driver's thinking time is increased to 2.5 s, will the car stop in time?

c) What happens if the thinking time is 1.5 s but the car is travelling at 64km / h

A boy walks to school. He walks 3 km in 30 minutes. He meets some friends and they talk for 10 minutes before they carry on walking to school. They walk 1 km in 15 minutes.

b) What was the average velocity of the boy's journey? Give your answer in m/s.

a) Draw a displacement-time graph to show the boy's journey to school.

1. Construct a glossary of the key terms in this unit. You could

add it to the one you made for Units 1 and 2. 2. Describe what happens to a ball when you drop it from a height

of 2 metres.

3. Explain the difference between average velocity and instantaneous velocity.

4. A bus travels 80 km due south in 2 hours. It then travels 100 km due west in 3 hours. What is the average velocity of the bus?

5. A car is travelling at 50km / h The driver sees a child run out into the road 5 m ahead. She applies the breaks and the car stops in 5 seconds. The driver's thinking time is 1.5 s.

a) Will the car stop in time?

b) If the driver's thinking time is increased to 2.5 s, will the car stop in time?

c) What happens if the thinking time is 1.5 s but the car is travelling at 64km / h

6. What assumption do you have to make if you are asked to do a

calculation on a falling body?

7. A boy walks to school. He walks 3 km in 30 minutes. He meets some friends and they talk for 10 minutes before they carry on walking to school. They walk 1 km in 15 minutes.

a) Draw a displacement-time graph to show the boy's journey to school.

b) What was the average velocity of the boy's journey? Give your answer in m / s

8. Explain, in terms of forces and acceleration, what happens

when a body is moving in uniform horizontal circular motion. 9. How do the forces on a body moving in a vertical circle vary?

10. What is the difference between radial and tangential acceleration?

11. What is relative velocity?

homework dump?

To determine whether the car will stop in time, we need to calculate the stopping distance and compare it to the distance to the child.

1. First, let's calculate the stopping distance:
Stopping distance = reaction distance + braking distance

a) The reaction distance is the distance the car travels during the driver's thinking time. Given that the driver's thinking time is 1.5 seconds, we can calculate the reaction distance using the formula: reaction distance = speed x time.
Given that the speed is 50 km/h (or 13.9 m/s) and the time is 1.5 seconds, the reaction distance is:
reaction distance = 13.9 m/s * 1.5 s = 20.85 meters.

b) If the driver's thinking time is increased to 2.5 seconds, the new reaction distance would be:
reaction distance = 13.9 m/s * 2.5 s = 34.75 meters.

Now, let's calculate the braking distance. The braking distance is the distance the car travels while decelerating to a stop.
The formula for braking distance is: braking distance = (speed^2) / (2 x acceleration).

c) If the car is traveling at 64 km/h (or 17.8 m/s), we need to calculate the braking distance using the formula above. However, we don't have the acceleration value. Assuming the car's deceleration is constant, we need to determine the acceleration first.

To find the acceleration, we can use the formula:
acceleration = (final speed - initial speed) / time

Given that the final speed is 0 m/s, the initial speed is 17.8 m/s, and the time is 5 seconds, we can calculate the acceleration:
acceleration = (0 m/s - 17.8 m/s) / 5 s = -3.56 m/s^2 (Note the negative sign indicates deceleration).

Now, we can calculate the braking distance using the formula:
braking distance = (speed^2) / (2 x acceleration)
braking distance = (17.8 m/s)^2 / (2 x -3.56 m/s^2)
braking distance = 158.44 m^2/s^2 / -7.12 m/s^2 = -22.23 meters (Note the negative sign indicates opposing direction).

Now let's compare the total stopping distance with the distance to the child:

a) The total stopping distance is the sum of the reaction distance and braking distance:
total stopping distance = reaction distance + braking distance
total stopping distance = 20.85 m + (-22.23 m) = -1.38 meters

Since the total stopping distance is negative, it means that the car would not stop in time to avoid hitting the child.

b) If the driver's thinking time is increased to 2.5 seconds, the new total stopping distance would be:
total stopping distance = reaction distance + braking distance
total stopping distance = 34.75 m + (-22.23 m) = 12.52 meters

In this case, since the total stopping distance is positive, the car would stop in time.

c) If the thinking time remains at 1.5 seconds but the car is traveling at 64 km/h, the braking distance would be calculated as:
braking distance = (speed^2) / (2 x acceleration)
braking distance = (17.8 m/s)^2 / (2 x -3.56 m/s^2)
braking distance = 158.44 m^2/s^2 / -7.12 m/s^2 = -22.23 meters

So, the braking distance would still be -22.23 meters. In this case, the total stopping distance would be:
total stopping distance = reaction distance + braking distance
total stopping distance = 20.85 m + (-22.23 m) = -1.38 meters

Again, as the total stopping distance is negative, it means that the car would not stop in time to avoid hitting the child.