A car is stopped at a traffic light. It then travels along a straight road such that its distance from the light is given by&nbsp;x(t)=bt2−ct3&nbsp;, where&nbsp;b=2.40ms2&nbsp;and&nbsp;c=0.120ms3&nbsp;(a) Calculate the average velocity of the car for the time interval t = 0 to t = 10.0 s. (b) Calculate the instantaneous velocity of the car at t=0, t = 5.0 s, and t = 10.0 s. (c) How long after starting from rest is the car again at rest?

Question

A car is stopped at a traffic light. It then travels along a straight road such that its distance from the light is given by&amp;nbsp;x(t)=bt2−ct3&amp;nbsp;, where&amp;nbsp;b=2.40ms2&amp;nbsp;and&amp;nbsp;c=0.120ms3&amp;nbsp;(a) Calculate the average velocity of the car for the time interval t = 0 to t = 10.0 s. (b) Calculate the instantaneous velocity of the car at t=0, t = 5.0 s, and t = 10.0 s. (c) How long after starting from rest is the car again at rest?

Pulad1971 · Accepted Answer

Step1&amp;nbsp;(a)&amp;nbsp;The displacement divided by the time period during which it occurred yields the average velocity.At each time t, we will calculate the distance.x(0)=Zero&amp;nbsp;&amp;nbsp;m&amp;nbsp;x(10)=2.40(102)−120(103)=120m&amp;nbsp;vx,avg=△x△t&amp;nbsp;vx,avg=xf−xitf−ti&amp;nbsp;vx,avg=120−010−0&amp;nbsp;=12m/s&amp;nbsp;Step 2&amp;nbsp;The instantaneous velocity is the derviative of x with respect to t&amp;nbsp;vx=&amp;nbsp;dx&amp;nbsp;&amp;nbsp;dt&amp;nbsp;&amp;nbsp;The distance equation will be differentiated.vx=2bt−3ct2&amp;nbsp;&amp;nbsp;(Eq.1)&amp;nbsp;We will substitute for different values of t into (Eq.1)&amp;nbsp;At&amp;nbsp;&amp;nbsp;t=0s&amp;nbsp;vx=0&amp;nbsp;At&amp;nbsp;&amp;nbsp;t=5s&amp;nbsp;vx=2(2.40)(5)−3(0.120)(5)2&amp;nbsp;=15m/s&amp;nbsp;At&amp;nbsp;&amp;nbsp;t=10s&amp;nbsp;vx=2(2.40)(10)−3(0.120)(10)2&amp;nbsp;=12ms&amp;nbsp;Step 3&amp;nbsp;(c)&amp;nbsp;The car is at rest means that&amp;nbsp;vx=0ms&amp;nbsp;By equating Eq.1 by Zero, we can find at what time the car will be at rest again.&amp;nbsp;2bt−3ct2=0&amp;nbsp;t=2b3c&amp;nbsp;t=2(2.40)3(0.120)&amp;nbsp;t=13.33s&amp;nbsp;Result&amp;nbsp;(a)vx,avg=12ms&amp;nbsp;(b)vx=0ms,vx=15ms,vx=12ms&amp;nbsp;(c)t=13.33s

RizerMix · Accepted Answer

(a) The average velocity of the car for the time interval t=0 to t=10.0 s can be calculated by finding the displacement of the car during this time interval and dividing it by the duration of the interval. The displacement of the car is given by:Δx=x(10.0)−x(0)=b(10.0)2−c(10.0)3−(0)2+c(0)3=240−0=240&amp;nbsp;mThe duration of the interval is 10.0 s. Therefore, the average velocity of the car during this interval is:vavg=ΔxΔt=24010.0=24&amp;nbsp;m/s(b) The instantaneous velocity of the car at t=0, t=5.0 s, and t=10.0 s can be calculated by taking the derivative of the distance function x(t) with respect to time t:v(t)=dxdt=2bt−3ct2At t=0, the instantaneous velocity is:v(0)=2b(0)−3c(0)2=0At t=5.0 s, the instantaneous velocity is:v(5.0)=2b(5.0)−3c(5.0)2=50−37.5=12.5&amp;nbsp;m/sAt t=10.0 s, the instantaneous velocity is:v(10.0)=2b(10.0)−3c(10.0)2=200−120=80&amp;nbsp;m/s(c) To find the time it takes for the car to come to rest, we need to find the time at which the velocity of the car becomes zero. Setting the velocity function v(t) to zero and solving for t, we get:2bt−3ct2=0t(2b−3ct)=0t=0 or t=2b3c=2(2.40)3(0.120)=40&amp;nbsp;sSince t=0 corresponds to the initial position of the car, we can conclude that the car comes to rest again at t=40 s after starting from rest.

Jeffrey Jordon · Accepted Answer

(a) We can find the average velocity of the car over the time interval t=0 to t=10.0 s by using the formula:vavg=total&amp;nbsp;distancetotal&amp;nbsp;timeThe total distance traveled by the car during this interval is the difference between the final position x(10.0) and the initial position x(0):total&amp;nbsp;distance=x(10.0)−x(0)=(2.40)(10.0)2−(0.120)(10.0)3=240&amp;nbsp;mThe total time elapsed is 10.0 s. Therefore, the average velocity is:vavg=240&amp;nbsp;m10.0&amp;nbsp;s=24&amp;nbsp;m/s(b) We can find the instantaneous velocity of the car at t=0, t=5.0 s, and t=10.0 s by taking the derivative of the distance function x(t) with respect to time t, as before. Alternatively, we can use the fact that the instantaneous velocity is equal to the slope of the tangent line to the distance function at the given time. The equation of the tangent line at time t is:y−x(t)=x′(t0)(t−t0)where y is the distance traveled by the car at time t, x(t) is the given distance function, x′(t0) is the derivative of the distance function evaluated at a nearby time t0 close to t, and t0 is the time at which the tangent line intersects the distance function.For example, at t=5.0 s, we have:y−x(5.0)=x′(4.9)(5.0−4.9)where x′(4.9) is the derivative of the distance function evaluated at t=4.9 s:x′(4.9)=2b(4.9)−3c(4.9)2=48.6&amp;nbsp;m/sSubstituting in the values, we get:y−117=48.6(0.1)y=117+4.86=121.86&amp;nbsp;m/sTherefore, the instantaneous velocity of the car at t=5.0 s is:v(5.0)=x′(5.0)=2b(5.0)−3c(5.0)2=12.5&amp;nbsp;m/sSimilarly, we can find the instantaneous velocity at t=0 and t=10.0 s.(c) To find the time it takes for the car to come to rest, we need to solve the equation 2bt−3ct2=0 for t. We can factor out t to get:t(2b−3ct)=0So either t=0 or 2b−3ct=0. Solving for t in the second equation, we get:t=2b3c=40&amp;nbsp;sTherefore, the car comes to rest again at t=40

A car is stopped at a traffic light. It then travels along a straight road such

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