Question

Two cars A and B, travel in a straight line. The distance of A from the starting point is given as a function of time by x????(????) = ???????? + ???????? 2 with ???? = 2.6 m ???? and ???? = 1.20 m/???? 2 . The distance of B from the starting point is x????(????) = ???????? 2 − ???????? 3 with ???? = 2.80 m/???? 2 and ???? = 0.2 m/???? 3 . [a].Which car is ahead just after they leave the starting point? [b].At what time(s) are the cars at the same point? [c]. At what time(s) is the distance from to neither increasing nor decreasing? [d].At what time(s) do A and B have the same acceleration?

Answer 1

Answer:

the Question is incomplete,the complete question is

"Cars A and B travel in a straight line. The distance of A from the starting point is given as a function of time by xA(t)=αt+βt2, with α=2.60m/s and β=1.20m/s2. The distance of B from the starting point is xB(t)=γt2−δt3, with y=2.80m/s2 and δ=0.20m/s3. (a) Which car is ahead just after the two cars leave the starting point? (b) At what time(s) are the cars at the same point? (c) At what time(s) is the distance from A to B neither increasing nor decreasing? (d) At what time(s) do A and B have the same acceleration?"

a.Car A is ahead after leaving the starting point

b.t=0secs,t=2.27secs and t=5.73secs

c.t=1secs and t=4.33secs

d .t=2.67secs

Explanation:

First we need to write out the giving expressions for the distance from the question

For distance A we have

$x_{a}(t)=\alpha t +\beta t^{2}$

and for distance B we have

$x_{b}(t)=\gamma t^{2} -\δt^{3}$.

a.  Note for us to compare the car that will be ahead we need to determine the initial velocity of each car i.e at t=0.

Since the expression for the velocity is not giving, we apply our knowledge of instantaneous motion which involves us differentiating the distance in order to get the velocity.

$V=\frac{dx}{dt}$

and recall  that for $y=ax^{n} \\\frac{dy}{dx}=a*nx^{n-1}$

for velocity of car  A, we have

$v_{a}=\frac{d(\alpha t +\beta t^{2})}{dt}\\v_{a}=\alpha +2\beta t$

we substitute t=o

$v_{a}=\alpha \\\\v_{a}=2.6m/s$

for velocity of car  B, we have

$v_{b}=\frac{d(\gamma t^{2} -\delta t^{3})}{dt}\\v_{b}= 2\gamma t -3\delta t^{2}$

at t=0

$v_{b}=0$

in conclusion, since the initial velocity of car An is greater than that of car B, Car A is ahead after leaving the starting point

b. for the time at which the cars are the same point, we need to find the time at which the cars have equal position. this can be done by  equating the position of both cars i.e

$x_{a}(t)=x_{b}(t)\\ \alpha t +\beta t^{2}=\gamma t^{2} -\δt^{3}$

if we carry out simply arithmetic, we arrive at

$\delta t^{2}+(\beta -\gamma)t +\alpha =0$

applying quadratic formula, we determine the value of t to be

$t=2.27secs \\t=5.73secs$

also note that at the starting point before the two cares take-off they are also at the same position i.e at t=0 hence the time which the cars are at the same point are

$t=0secs\\t=2.27secs \\t=5.73secs$

c. the distance will be constant when the two cars move at the same velocity i.e

$v_{a}=v_{b}$

$\alpha +2\beta t=2\gamma t -3\delta t^{2}$

if we rearrange the equation and substitute the value of the constants, we arrive at

$0.6t^{2} -3.2t +2.6=0$

Solving the simple quadratic equation, we arrive at the values of t which are

$t=1secs\\t=4.33secs$

d. To determine the time at which the two cars will have the same acceleration, we need to determine the expression for the acceleration which we can get by differentiating the the velocity for each car

$a_{a}(t)=\frac{d(\alpha +2\beta t) }{dt}\\ a_{a}(t)=2\beta$

and for car B we have the acceleration to be

$a_{b}(t)=\frac{d(2\gamma t - 3\delta t^{2})}{dt}\\ a_{b}(t)=2\gamma -6\delta t$

Hence to have the same acceleration

$a_{a}(t)=a_{b}(t)\\2\beta=2\gamma -6\delta t$

Making t the subject of formula,

$t=\frac{\gamma - \beta}{3\delta}\\t=\frac{2.8-1.3}{3*0.2}\\ t=2.67secs$

Hence the cars will have the same acceleration at t=2.67secs

Answer 2

Answer:

a) They are in the same point

b) t = 0 s, t = 2.27 s, t = 5.73 s

c) t = 1 s, t = 4.33 s

d) t = 2.67 s

Explanation:

Given equations are:

$x_{a}(t) = at+bt^2$

$x_{b}(t) = ct^2-dt^3$

Constants are:

$a = 2.60 m/s, b = 1.20 m/s^2, c= 2.80 m/s^2, d = 0.20 m/s^3$

a) "Just after leaving the starting point" means that t = 0. So, if we look the equations, both $x_a(t)$ and $x_b(t)$ depend on t and don't have constant terms.

So both cars A and B are in the same point.

b) Firstly, they are in the same point in x = 0 at t = 0. But for generalized case, we must equalize equations and solve quadratic equation where roots will give us proper t value(s).

$at+bt^2 = ct^2-dt^3$

$2.6t + 1.2t^2 = 2.8t^2 - 0.2t^3\\0.2t^2 - 1.6t + 2.6 = 0\\t^2 - 8t + 13 = 0$

$t_1 = 4 - \sqrt{3} = 2.27$ s, $t_1 = 4 + \sqrt{3} = 5.73$ s

c) Since the distance isn't changing, the velocities are equal. To find velocities, we need to take the derivatives of both equations with respect to time and equalize them.

$v_a(t) = \frac{d}{d(t)}x_a(t) = a + 2bt \\v_b(t) = \frac{d}{d(t)}x_b(t) = 2ct - 3dt^2\\a+2bt = 2ct - 3dt^2\\3dt^2+2(b-c)t+a = 0\\0.6t^2-3.2t+2.6 = 0$

$t_1 = 1$ s, $t_2 = 4.33$ s

d) For same acceleration, we we need to take the derivatives of velocity equations with respect to time and equalize them.

$a_a(t) = \frac{d}{d(t)}v_a(t) = 2b \\a_b(t) = \frac{d}{d(t)}v_b(t) = 2c - 6dt\\2b = 2c - 6dt\\3dt = c - b\\t = (c - b)/3d = (2.8 - 1.2)/(3*0.2) = 2.67$ s