A car starts from rest and travels for 5.2 s with a uniform acceleration of +1.6 m/s2. The driver then applies the brakes, causing a uniform acceleration of -2.1 m/s2. The breaks are applied for 1.50 s.
(a) How far has the car gone from its start?
speed when brakes applied ... 5.2 s * 1.6 m/s^2 = 8.3 m/s
speed after braking ... 8.3 m/s + (1.50 s * -2.1 m/s^2) = 5.2 m/s
use the times and average speeds over the two segments
... to find the distance traveled
Thank you, but I tried doing that and got an answer of 90.45 and said that was wrong... I'm not sure what I'm doing wrong.
check your math
the car is moving for a total of 6.7 s ... 90.45 is too high
also, watch the sig figs ... probably a max of one decimal place in the answer
To find the distance the car has traveled from its start, we need to calculate the distance covered during each phase of motion (acceleration and deceleration) and then add them together.
First, let's calculate the distance covered during the acceleration phase. We will use the equation:
s = ut + (1/2)at^2
Where:
s = distance
u = initial velocity (which is 0 m/s since the car starts from rest)
a = acceleration
t = time
Given:
u = 0 m/s
a = +1.6 m/s^2
t = 5.2 s
Using the formula with the given values, we have:
s1 = 0 + (1/2)(1.6)(5.2)^2
Calculating this, we get:
s1 = 0 + 0.5 * 1.6 * 27.04
s1 = 0 + 21.824
s1 = 21.824 m
Now, let's calculate the distance covered during the deceleration (braking) phase. Again, we will use the same formula, but this time with a negative acceleration:
s = ut + (1/2)at^2
Where:
s = distance
u = initial velocity (which is the final velocity at the end of the acceleration phase)
a = acceleration
t = time
To find the initial velocity for the deceleration phase, we need to calculate the final velocity from the acceleration phase. We can use the equation:
v = u + at
Where:
v = final velocity
u = initial velocity (which is 0 m/s)
a = acceleration
t = time
Given:
u = 0 m/s
a = +1.6 m/s^2
t = 5.2 s
Using the formula with the given values, we have:
v = 0 + (1.6)(5.2)
v = 0 + 8.32
v = 8.32 m/s
Now, let's calculate the distance during the deceleration phase. We have:
s2 = 8.32 * 1.5 + (1/2)(-2.1)(1.5)^2
Calculating this, we get:
s2 = 12.48 + (-0.5 * 2.1 * 2.25)
s2 = 12.48 + (-2.205)
s2 = 12.48 - 2.205
s2 = 10.275 m
Finally, we can calculate the total distance traveled by adding the distance covered during the acceleration and deceleration phases:
Total distance = s1 + s2
Total distance = 21.824 + 10.275
Total distance = 32.099 m
Therefore, the car has traveled approximately 32.099 meters from its start.