An object accelerates, uniformly from rest. If it travels 26 m south, and reaches a velocity of 11 m/s south, how long was the object accelerating?
Answers
Answer:
A motorboat is traveling at a constant velocity of 5.0 m/s when it starts to decelerate to arrive at the dock. Its acceleration is
\[a(t)=-\frac{1}{4}t\,\text{m/}{\text{s}}^{2}\]
. (a) What is the velocity function of the motorboat? (b) At what time does the velocity reach zero? (c) What is the position function of the motorboat? (d) What is the displacement of the motorboat from the time it begins to decelerate to when the velocity is zero? (e) Graph the velocity and position functions.
Strategy
(a) To get the velocity function we must integrate and use initial conditions to find the constant of integration. (b) We set the velocity function equal to zero and solve for t. (c) Similarly, we must integrate to find the position function and use initial conditions to find the constant of integration. (d) Since the initial position is taken to be zero, we only have to evaluate the position function at
\[t=0\]
.
Solution
We take t = 0 to be the time when the boat starts to decelerate.
From the functional form of the acceleration we can solve (Figure) to get v(t):
[reveal-answer q=”136447″]Show Answer[/reveal-answer]
[hidden-answer a=”136447″]
\[v(t)=\int a(t)dt+{C}_{1}=\int -\frac{1}{4}tdt+{C}_{1}=-\frac{1}{8}{t}^{2}+{C}_{1}.\]
At t = 0 we have v(0) = 5.0 m/s = 0 + C1, so C1 = 5.0 m/s or
\[v(t)=5.0\,\text{m/}\text{s}-\frac{1}{8}{t}^{2}\]
.[/hidden-answer]
[reveal-answer q=”967265″]Show Answer[/reveal-answer]
[hidden-answer a=”967265″]
\[v(t)=0=5.0\,\text{m/}\text{s}-\frac{1}{8}{t}^{2}⇒t=6.3\,\text{s}\]
[/hidden-answer]
Solve (Figure):
[reveal-answer q=”251505″]Show Answer[/reveal-answer]
[hidden-answer a=”251505″]
\[x(t)=\int v(t)dt+{C}_{2}=\int (5.0-\frac{1}{8}{t}^{2})dt+{C}_{2}=5.0t-\frac{1}{24}{t}^{3}+{C}_{2}.\]
At t = 0, we set x(0) = 0 = x0, since we are only interested in the displacement from when the boat starts to decelerate. We have
\[x(0)=0={C}_{2}.\]
Therefore, the equation for the position is
\[x(t)=5.0t-\frac{1}{24}{t}^{3}.\]
[/hidden-answer]
[reveal-answer q=”330950″]Show Answer[/reveal-answer]
[hidden-answer a=”330950″]Since the initial position is taken to be zero, we only have to evaluate x(t) when the velocity is zero. This occurs at t = 6.3 s. Therefore, the displacement is
Answer:
Explanation:
Here we have to find the time so,direction is given to confuse you
Given,
Distance=26m south(travelling)
Velocity=11m/s
Now,
VELOCITY=DISTANCE/TIME
substitute the values
11=26/t
t=2.3636s