A bus accelerates at 1.4 m/s2 from rest for 10 s. It then travels at constant speed for 26 s, after which it slows to a stop with an acceleration of magnitude 1.4 m/s2.
(a) What is the total distance that the bus travels?
(b) What was its average velocity?
distance=1/2*1.4*10^2 + 1.4*10*26+ gets you the distance till it starts to brake.
Now to deaccelerate from that velocity, it takes 10 sec (think on that).
distance braking= vi*t-1/2 a t^2 or
1.4*10*10-1/2 1.4 10^2=70m
add that to the above distance.
avg velocity= distance/total time=totaldistance/(10+26+10)
To find the total distance that the bus travels, we need to calculate the distance traveled during each phase of its motion.
(a) Distance during acceleration phase:
During this phase, the bus is accelerating from rest at a constant rate of 1.4 m/s^2 for a duration of 10 seconds. We can use the formula for distance traveled during uniform acceleration:
d = (1/2) * a * t^2
where d is the distance, a is the acceleration, and t is the time.
Plugging in the given values, we have:
d = (1/2) * 1.4 * (10)^2
= 0.5 * 1.4 * 100
= 70 meters
So, the distance traveled during the acceleration phase is 70 meters.
Distance during constant speed phase:
The bus travels at a constant speed for 26 seconds. During this phase, the bus does not change its position since its speed is constant. Therefore, the distance traveled is given by:
distance = speed * time
= constant speed * 26
However, we don't have the value for the constant speed of the bus. So, we can't calculate the distance during this phase.
Distance during deceleration phase:
The bus then slows to a stop with an acceleration of magnitude 1.4 m/s^2. Since the bus is coming to a stop, we can ignore the time it takes to decelerate because the final velocity is zero.
We can use the formula for calculating the distance traveled during uniform acceleration, where the final velocity is zero:
d = (v^2 - u^2) / (2 * a)
where v is the final velocity, u is the initial velocity (which is the constant speed of the bus), and a is the acceleration.
In this case, v is 0, u is the unknown constant speed of the bus during the constant speed phase, and a is -1.4 m/s^2 (negative because the bus is decelerating).
Therefore,
0 = (u^2 - 0^2) / (2 * -1.4)
u^2 = 0
u = 0
Since the initial velocity is also zero, the distance traveled during the deceleration phase is also zero.
Total distance traveled:
To find the total distance traveled, we sum up the distances traveled during each phase:
total distance = distance during acceleration phase + distance during constant speed phase + distance during deceleration phase
= 70 + constant speed * 26 + 0
= 70 + constant speed * 26
Note: Without knowing the constant speed during the constant speed phase, we cannot calculate the total distance traveled.
(b) To find the average velocity, we need to divide the total displacement (change in position) by the total time taken.
Since the bus started and ended at the same position, the total displacement is zero (there is no net change in position).
Therefore, the average velocity of the bus will also be zero.
In summary:
(a) The total distance traveled is 70 + constant speed * 26 meters, where the constant speed during the constant speed phase is not given, so it cannot be determined.
(b) The average velocity of the bus is zero.
To find the total distance that the bus travels, we need to break down the problem into three parts - the initial acceleration, the constant speed, and the deceleration.
(a) Distance during the initial acceleration:
We can use the formula for calculating distance with constant acceleration:
\[d = \frac{1}{2} \cdot a \cdot t^2\]
where \(d\) is the distance, \(a\) is the acceleration, and \(t\) is the time.
Given:
\(a = 1.4 \, \text{m/s}^2\)
\(t = 10 \, \text{s}\)
Substituting the values into the equation, we get:
\(d_{\text{acceleration}} = \frac{1}{2} \cdot 1.4 \, \text{m/s}^2 \cdot (10 \, \text{s})^2\)
\(d_{\text{acceleration}} = 70 \, \text{m}\)
(b) Distance during the constant speed:
Since the bus is traveling at a constant speed, the distance is expressed as:
\[d_{\text{constant speed}} = v \cdot t\]
where \(v\) is the velocity and \(t\) is the time.
Given:
\(v = \text{constant}\)
\(t = 26 \, \text{s}\)
Substituting the values, we get:
\(d_{\text{constant speed}} = v \cdot 26 \, \text{s}\)
(c) Distance during deceleration:
Using the same formula as in part (a):
\[d = \frac{1}{2} \cdot a \cdot t^2\]
Given:
\(a = -1.4 \, \text{m/s}^2\) (negative sign indicates deceleration)
\(t = \text{time taken to decelerate to zero velocity}\)
To find the time taken to decelerate to zero velocity, we can use the formula:
\(v = u + a \cdot t\)
where \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time.
Since the final velocity \(v\) is zero, we have:
\(0 = u + (-1.4 \, \text{m/s}^2) \cdot t\)
Solving for \(t\), we get:
\(t = \frac{u}{1.4 \, \text{m/s}^2}\)
Now we can substitute the value of \(t\) into the distance formula:
\(d_{\text{deceleration}} = \frac{1}{2} \cdot (-1.4 \, \text{m/s}^2) \cdot \left(\frac{u}{1.4 \, \text{m/s}^2}\right)^2\)
To find \(u\), the initial velocity during the constant speed phase, we can use the fact that velocity is the rate of change of distance:
\[u = \frac{d_{\text{constant speed}}}{t_{\text{constant speed}}}\]
Given that \(t_{\text{constant speed}} = 26 \, \text{s}\), we can substitute the values into the equation and solve for \(u\).
Finally, the total distance traveled by the bus is:
\[d_{\text{total}} = d_{\text{acceleration}} + d_{\text{constant speed}} + d_{\text{deceleration}}\]
(b) To find the average velocity, we can use the formula:
\[v_{\text{average}} = \frac{\text{total distance}}{\text{total time}}\]
The total time can be calculated by adding the times for each phase:
\[t_{\text{total}} = t_{\text{acceleration}} + t_{\text{constant speed}} + t_{\text{deceleration}}\]
Given that \(t_{\text{acceleration}} = 10 \, \text{s}\), \(t_{\text{constant speed}} = 26 \, \text{s}\), and \(t_{\text{deceleration}}\) is the time calculated earlier for deceleration.
Substituting the values into the equation, we can find \(v_{\text{average}}\).
It is important to note that calculating \(u\) and \(t_{\text{deceleration}}\) depends on the specific information given in the problem, such as the initial velocity during the constant speed phase and the deceleration time.