A car starts from rest and travels for 5.0s with a uniform acceleration of 1.5m/s^2. The driver then applies the brakes, causing a uniform negative accelreeation of -2.0m/s^2. If the brakes are applied for 3.0s, how fast is the car going at the end of the braking period, and how far has it gone?

Answer in back of the book: 1.50m/s, 32.3m

do u now how to solve it ??

To solve this problem, we can use the equations of motion. Let's break down the different parts of the car's motion.

First, the car starts from rest and travels for 5.0 seconds with a uniform acceleration of 1.5 m/s^2. We can use the equation:

v = u + at

Where:
v = final velocity (what we are looking for)
u = initial velocity (0 m/s, as the car starts from rest)
a = acceleration (1.5 m/s^2)
t = time (5.0 seconds)

Plugging in the values:

v = 0 + (1.5)(5.0)
v = 7.5 m/s

So at the end of the initial acceleration period, the car's velocity is 7.5 m/s.

Next, the driver applies the brakes, causing a uniform negative acceleration of -2.0 m/s^2. The time for braking is given as 3.0 seconds. Again, we can use the equation:

v = u + at

This time, the initial velocity (u) is the final velocity (7.5 m/s) from the previous step. The acceleration (a) is -2.0 m/s^2, and the time (t) is 3.0 seconds.

v = 7.5 + (-2.0)(3.0)
v = 7.5 - 6.0
v = 1.5 m/s

So at the end of the braking period, the car's velocity is 1.5 m/s.

To determine the distance traveled during the entire motion, we can use the equation:

s = ut + 1/2at^2

For the first part of the motion (uniform acceleration):

s1 = (0)(5.0) + 1/2(1.5)(5.0)^2
s1 = 0 + 1/2(1.5)(25.0)
s1 = 18.75 m

For the second part of the motion (uniform negative acceleration):

s2 = (7.5)(3.0) + 1/2(-2.0)(3.0)^2
s2 = 22.5 + 1/2(-2.0)(9.0)
s2 = 22.5 + (-9.0)
s2 = 13.5 m

To find the total distance traveled, we sum up s1 and s2:

s_total = s1 + s2
s_total = 18.75 + 13.5
s_total = 32.25 m

From the calculations, we find that the car is traveling at a speed of 1.5 m/s at the end of the braking period and has traveled a total distance of 32.25 m.