A car starts from rest and travels for 5.9 s with a uniform acceleration of +1.3 m/s2. The driver then applies the brakes, causing a uniform acceleration of -1.7 m/s2. The breaks are applied for 1.60 s.

(a) How fast is the car going at the end of the braking period?
2.6
m/s
(b) How far has the car gone from its start?
M

To solve this problem, we can use the equations of motion. Let's break down the problem into two parts: the acceleration phase and the braking phase.

(a) To find the final velocity at the end of the braking period, we can use the equation:

v = u + at

where:
v = final velocity
u = initial velocity (which is the velocity at the end of the acceleration phase)
a = acceleration during the braking phase
t = time taken during the braking phase

During the acceleration phase:
Initial velocity (u) = 0 m/s (the car starts from rest)
Time (t) = 5.9 s
Acceleration (a) = +1.3 m/s^2

v = u + at
v = 0 m/s + (1.3 m/s^2)(5.9 s)
v = 7.67 m/s

Therefore, the car is going at a speed of 7.67 m/s at the end of the braking period.

(b) To find the distance traveled from the start, we can use the equation:

s = ut + (1/2)at^2

During the acceleration phase:
Initial velocity (u) = 0 m/s
Time (t) = 5.9 s
Acceleration (a) = +1.3 m/s^2

s = (0 m/s)(5.9 s) + (1/2)(1.3 m/s^2)(5.9 s)^2
s = 0 + (0.5)(1.3 m/s^2)(34.81 s^2)
s = 22.64 m

During the braking phase:
Initial velocity (u) = 7.67 m/s (from part (a))
Time (t) = 1.60 s
Acceleration (a) = -1.7 m/s^2

s = (7.67 m/s)(1.60 s) + (1/2)(-1.7 m/s^2)(1.60 s)^2
s = 12.27 m - 2.73 m
s = 9.54 m

Therefore, the car has gone a total distance of 22.64 m + 9.54 m = 32.18 m from its start.

To get the answer to this question, we can use the equations of kinematics.

First, let's find the final velocity of the car after the acceleration period. We can use the equation:

v = u + at

where v is the final velocity, u is the initial velocity (which is 0 since the car starts from rest), a is the acceleration (+1.3 m/s^2 in this case), and t is the time (5.9 s).

Plugging in the values, we get:

v = 0 + (1.3 m/s^2) * (5.9 s)
v = 7.67 m/s

So, at the end of the acceleration period, the car is going at a speed of 7.67 m/s.

Now, let's find the final velocity of the car after the braking period. Again, we can use the same equation of motion:

v = u + at

where v is the final velocity, u is the initial velocity (which is the velocity acquired after the acceleration period, 7.67 m/s), a is the acceleration (-1.7 m/s^2 this time), and t is the time of braking (1.60 s).

Plugging in the values, we get:

v = 7.67 m/s + (-1.7 m/s^2) * (1.60 s)
v = 7.67 m/s - 2.72 m/s
v = 4.95 m/s

So, at the end of the braking period, the car is going at a speed of 4.95 m/s.

To find how far the car has gone from its start, we can use the equation:

s = ut + (1/2)at^2

where s is the displacement (distance traveled), u is the initial velocity (0), a is the acceleration (-1.7 m/s^2), and t is the time of braking (1.60 s).

Plugging in the values, we get:

s = 0 + (1/2)(-1.7 m/s^2)(1.60 s)^2
s = -(1.7 m/s^2)(2.56 s^2)
s = -4.35 m

The negative sign indicates that the car has traveled in the opposite direction. So, the car has gone 4.35 meters backward from its start.

Therefore, the answers to the questions are:
(a) The car is going at a speed of 4.95 m/s at the end of the braking period.
(b) The car has traveled 4.35 meters backward from its start.

a. d1 = 0.5a*t^2 = 0.5*1.3*5.9^2=22.63 m

V1 = a*t = 1.3 * 5.9 = 7.67 m/s.

V = V1 + a*t = 7.67 - 1.7*1.6 = 4.95 m/s

b. d1 = 0.5a*t^2 = 0.5*1.3*5.9^2=22.63 m

d2 = (V^2-V1^2/2a =
(4.95^2-(7.67^2))/-3.4 = 10.1 m.

d = d1+d2 = 22.63 + 10.1 = 32.73 m