A car starts from rest and travels for 5.3 s with a uniform acceleration of +1.2 m/s2. The driver then applies the brakes, causing a uniform acceleration of -2.3 m/s2. The breaks are applied for 1.60 s.

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

How far has the car gone from its start?

Well, it seems like this car can't make up its mind whether it wants to speed up or slow down! But let's calculate the answers and see where it ends up, shall we?

To find the final velocity after the braking period, we can use the equation:

v = u + a * t

Where:
v = final velocity
u = initial velocity (in this case 0 m/s since the car starts from rest)
a = acceleration (in this case -2.3 m/s^2 since we're talking about the braking period)
t = time (in this case 1.60 s)

v = 0 + (-2.3) * 1.60

Calculating that gives us v = -3.68 m/s. So, the car is going backwards at 3.68 m/s. Guess it really doesn't like brakes!

Now, to find the distance traveled from the start, we can use the equation:

s = u * t + (1/2) * a * t^2

Where:
s = distance traveled
u = initial velocity (again 0 m/s)
t = time (in this case 5.3 s)
a = acceleration (in this case 1.2 m/s^2)

s = 0 * 5.3 + (1/2) * 1.2 * (5.3)^2

Calculating that gives us s = 8.994 m. So, the car has traveled approximately 8.994 meters. Not too far, but at least it's not stuck in the same spot!

Oh, the things we do for physics calculations. But remember, safety first! Always wear your seatbelt and follow the rules of the road.

To find the velocity of the car at the end of the braking period, we need to calculate the final velocity using the equation:

v = u + at

where:
v = final velocity
u = initial velocity (0 m/s since the car starts from rest)
a = acceleration during the braking period (-2.3 m/s^2)
t = time of the braking period (1.60 s)

Substituting the values into the equation:

v = 0 + (-2.3) x 1.60
v = -3.68 m/s

Therefore, the car is traveling at a speed of -3.68 m/s at the end of the braking period.

To calculate the distance the car has traveled from its start, we need to find the total displacement during the time of acceleration and the time of braking.

During the time of acceleration (5.3 s) with a uniform acceleration of +1.2 m/s^2, we can use the equation:

s = ut + 0.5at^2

where:
s = displacement
u = initial velocity (0 m/s since the car starts from rest)
a = acceleration during the acceleration period (1.2 m/s^2)
t = time of the acceleration period (5.3 s)

Substituting the values into the equation:

s1 = 0 + 0.5(1.2)(5.3)^2
s1 = 17.97 m

During the time of braking (1.60 s) with a uniform acceleration of -2.3 m/s^2, we use the same equation:

s = ut + 0.5at^2

where:
s = displacement
u = initial velocity (0 m/s since the car starts from rest)
a = acceleration during the braking period (-2.3 m/s^2)
t = time of the braking period (1.60 s)

Substituting the values into the equation:

s2 = 0 + 0.5(-2.3)(1.60)^2
s2 = -2.94 m

The negative sign indicates that the displacement during the braking period is in the opposite direction (opposite of the initial direction of motion).

To find the total displacement from the start, we add the displacement during the acceleration and the displacement during the braking:

total displacement = s1 + s2
total displacement = 17.97 + (-2.94)
total displacement = 15.03 m

Therefore, the car has traveled a total distance of 15.03 m from its start.

To find the car's velocity at the end of the braking period, we need to determine its velocity after the first period of acceleration and then calculate the change in velocity caused by the braking.

First, let's find the car's velocity after the initial period of acceleration. We know that the car starts from rest and accelerates uniformly for 5.3 seconds with an acceleration of +1.2 m/s². We can use the equation:

v = u + at

Where:
v = final velocity
u = initial velocity (0 m/s as the car starts from rest)
a = acceleration (+1.2 m/s²)
t = time (5.3 s)

Plugging in the values, we get:

v = 0 + (1.2 m/s²)(5.3 s)
v = 6.36 m/s

Therefore, the car's velocity after the initial acceleration is 6.36 m/s.

Now, we need to find the change in velocity due to the braking period. The car experiences a uniform acceleration of -2.3 m/s² for 1.60 seconds. We can again use the same equation:

v = u + at

Where:
v = final velocity at the end of the braking period
u = initial velocity (6.36 m/s from the previous calculation)
a = acceleration (-2.3 m/s²)
t = time (1.60 s)

Plugging in the values, we get:

v = 6.36 m/s + (-2.3 m/s²)(1.60 s)
v = 6.36 m/s - 3.68 m/s
v = 2.68 m/s

Therefore, the car's velocity at the end of the braking period is 2.68 m/s.

To find how far the car has traveled from its start, we can use another equation that relates initial velocity, final velocity, acceleration, and displacement:

s = ut + (1/2)at²

Where:
s = displacement
u = initial velocity (0 m/s as the car starts from rest)
t = time

For the first period of acceleration:
s₁ = 0 + (1/2) (1.2 m/s²) (5.3 s)²
s₁ = 0 + (1/2) (1.2 m/s²) (28.09 s²)
s₁ = 0 + 16.854 m
s₁ ≈ 16.854 m

For the braking period:
s₂ = 6.36 m/s (1.60 s) + (1/2) (-2.3 m/s²) (1.60 s)²
s₂ = 10.176 m - 7.3936 m
s₂ = 2.7824 m

The total distance traveled by the car is the sum of these two distances:

Total distance = s₁ + s₂
Total distance = 16.854 m + 2.7824 m
Total distance ≈ 19.6364 m

Therefore, the car has traveled approximately 19.6364 meters from its start.

d1 = 0.5a*t^2 = 0.6*5.3^2 = 16.85 m.

V1 = a*t = 1.2m/s^2 * 5.3s = 6.36 m/s.

a. V = Vo + a.t
V = 6.36m/s - 2.3m/s^2*1.60s = 2.68 m/s.

b. V^2 = V0^2 + 2a*d
d2 = V^2-Vo^2)/2a
d2 = (2.68^2-6.36^2)/-4.6 = 7.23 m.

D = d1 + d2 = 16.85 + 7.23 = 24.1 m.