If a car starts from rest and travels for 5.0 seconds with a uniform acceleration of +1.5 m/s2. The driver then applies the brakes for 3.0 seconds, causing a uniform acceleration of - 2.1 m/s2.

a) Sketch the velocity vs. time graph for the car and paste your sketch in the space below.

b) How fast is the car going at the end of the braking period?

c) How far has it gone at the end of the braking period?

d) Name the three controls on an automobile which cause acceleration

a) I cant sketch it. Uniform acceleration means a linear increase of velocity (slope=acceleration). brakes mean a decrease in velocity, again, slope= acceleration.

b.calculate it.
c. distance= average velocity*time
d. steering wheel, accelerator, brakes,

A) v=u+at

a) As an AI text-based interface, I am unable to sketch the graph. However, I can describe the velocity vs. time graph for the car:

- For the initial 5.0 seconds with a uniform acceleration of +1.5 m/s^2, the velocity will increase linearly with time.
- After 5.0 seconds, the driver applies the brakes for 3.0 seconds, causing a uniform acceleration of -2.1 m/s^2. During this period, the velocity will decrease linearly with time.

b) To determine the car's speed at the end of the braking period, we need to calculate the final velocity. Let's consider the initial velocity as 0 m/s at rest.

For the first phase of 5.0 seconds with a uniform acceleration of +1.5 m/s^2:
v = u + at
v = 0 + (1.5 m/s^2) * 5.0 s
v = 7.5 m/s

For the second phase of 3.0 seconds with a uniform acceleration of -2.1 m/s^2:
v = u + at
v = 7.5 m/s + (-2.1 m/s^2) * 3.0 s
v = 7.5 m/s - 6.3 m/s
v = 1.2 m/s

Therefore, at the end of the braking period, the car's speed is 1.2 m/s.

c) To determine the distance traveled at the end of the braking period, we need to calculate the total distance covered in both phases.

For the first phase of 5.0 seconds with a uniform acceleration of +1.5 m/s^2:
s1 = ut + (1/2)at^2
s1 = (0 m/s) * 5.0 s + (1/2) * (1.5 m/s^2) * (5.0 s)^2
s1 = 0 m + (1/2) * 1.5 m/s^2 * 25.0 s^2
s1 = (1/2) * 1.5 m/s^2 * 25.0 s^2
s1 = 18.75 m

For the second phase of 3.0 seconds with a uniform acceleration of -2.1 m/s^2:
s2 = ut + (1/2)at^2
s2 = (7.5 m/s) * 3.0 s + (1/2) * (-2.1 m/s^2) * (3.0 s)^2
s2 = 22.5 m + (1/2) * (-2.1 m/s^2) * 9.0 s^2
s2 = 22.5 m - 9.45 m
s2 = 13.05 m

Therefore, at the end of the braking period, the car has traveled a total distance of 18.75 m + 13.05 m = 31.8 m.

d) The three controls on an automobile that cause acceleration are:

1. Accelerator pedal: Pressing down on the accelerator pedal increases the flow of fuel to the engine, resulting in an increase in engine power and acceleration of the vehicle.
2. Brake pedal: Pressing down on the brake pedal activates the braking system, which applies friction to slow down or stop the vehicle. The braking system causes deceleration or negative acceleration.
3. Clutch pedal (in manual transmission vehicles): Pressing down on the clutch pedal disengages the engine from the transmission, allowing the driver to shift gears. Proper gear shifting can affect the acceleration and performance of the vehicle.

a) To sketch the velocity vs. time graph for the car, we first need to determine the velocity at different points in time. We can use the equations of motion to do this.

For the first 5.0 seconds, the car experiences a uniform acceleration of +1.5 m/s². We can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Since the car starts from rest, u = 0. Plugging in the values, we have v = 0 + 1.5(5.0) = 7.5 m/s.

During the next 3.0 seconds, the car undergoes a uniform acceleration of -2.1 m/s². Again, using the same equation, we have v = u + at. The initial velocity at this point is 7.5 m/s (from the previous calculation). Plugging in the values, we have v = 7.5 + (-2.1)(3.0) = 1.2 m/s.

So, at the end of the braking period (t = 3.0 seconds), the car is traveling at a velocity of 1.2 m/s.

To sketch the velocity vs. time graph, we would draw a straight line with positive slope for the first 5.0 seconds (representing the acceleration), and then a straight line with negative slope for the next 3.0 seconds (representing the deceleration). The y-axis would represent velocity, while the x-axis would represent time.

b) The velocity of the car at the end of the braking period is 1.2 m/s.

c) To determine the distance traveled by the car at the end of the braking period, we can use the equation s = ut + (1/2)at², where s is the distance, u is the initial velocity, t is the time, and a is the acceleration.

For the first 5.0 seconds, the car still has the acceleration of +1.5 m/s². Since it starts from rest, u = 0. Plugging in the values, we have s = 0 + (1/2)(1.5)(5.0)² = 18.75 m.

During the next 3.0 seconds, the car decelerates with an acceleration of -2.1 m/s². The initial velocity at this point is 7.5 m/s (from the previous calculation). Plugging in the values, we have s = 7.5(3.0) + (1/2)(-2.1)(3.0)² = 22.95 m.

Therefore, at the end of the braking period (t = 3.0 seconds), the car has traveled a total distance of 18.75 + 22.95 = 41.70 m.

d) The three controls on an automobile that cause acceleration are:
1. Accelerator (Gas) Pedal: Pressing down the accelerator pedal increases the flow of fuel to the engine, which increases the engine power and ultimately accelerates the car.
2. Brake Pedal: Pressing down the brake pedal activates the braking system, which applies friction to the wheels and slows down or stops the car.
3. Clutch (if applicable): In manual transmission cars, disengaging the clutch by pressing the pedal interrupts the power transfer between the engine and the wheels, allowing the driver to shift gears and control the acceleration.