Physics, 18.12.2019 20:31 jakepeavy70

Starting from rest at the origin, x0 = 0, a car accelerates in a straight line along the +x direction. for the time interval from t0 = 0 s to t1= 4 s, its velocity is given by v(t) = (−10 m/s2 )t + (6 m/s3)t2 . then from t1 = 4 s to t2 = 10 s, it maintains a constant velocity v = 56 m/s in the same direction.

(i) find the average acceleration for the first time interval, from t0 = 0 s to t1 = 4 s.
(ii) find the average acceleration for the second time interval, from t1 = 4 s to t2 = 10 s
(iii) find the instantaneous acceleration a(t) as a function of time for 0 s < t < 4 s.
(iv) find the position x(t) as a function of time for 0 s < t < 4 s

Answers: 1

Show answers

Answers

Answer from: ella3714

i) The average acceleration for the first time interval, from $t_{0} = 0$ to $t_{1} =4\ s$ , is $14 \frac{m}{s^{2} }$ .

ii) The average acceleration for the second time interval, from $t_{1} =4\ s$ to $t_{2} =10\ s$ , is zero.

iii) The instantaneous acceleration as a function of time a(t) between 0 s and 4 s is $a(t)=-10\frac{m}{s^{2} } + 12\frac{m}{s^{3} } t$ .

iv) The position as a function of time x(t) between 0 s and 4 s is $x(t)=-5\frac{m}{s^{2} } t^{2} + 2\frac{m}{s^{3} } t^{3}$ .

Explanation:

We are given as data that a car that is a rest, that means v₀=0, at the origin, x₀=0, accelerates in a straight line along the +x direction. We are told that from $t_{0} = 0$ to $t_{1} =4\ s$ the velocity is $v(t)=-10\frac{m}{s^{2} } t + 6 \frac{m}{s^{3} } t^{2}$ and that from $t_{1} =4\ s$ to $t_{2} =10\ s$ it maintains a constant velocity of $v=56\frac{m}{s}$ in the same direction.

i) The average acceleration is simply the rate of change of velocity, it can be expressed as:

$\bar{a}=\frac{\vartriangle v}{\vartriangle t} =\frac{v_{f}-v_{0} }{t_{f}-t_{0}}$

we have to replace the values in this equation, the only value we need to calculate is $v_{f}$

$v(4 s)=-10\frac{m}{s^{2} } 4 s + 6 \frac{m}{s^{3} } (4 s)^{2}$

$v(4 s)=-10\frac{m}{s^{2} } 4 s + 6 \frac{m}{s^{3} } 16 s^{2}$

$v(4 s)= -40 \frac{m}{s} + 96 \frac{m}{s}= 56 \frac{m}{s}$

we get that $v_{f}= 56 \frac{m}{s}$ .

Then we pur this value in the expression of the average acceleration

$\bar{a}=\frac{\vartriangle v}{\vartriangle t} =\frac{v_{f}-v_{0}}{t_{f}-t_{0}}=\frac{56\frac{m}{s}-0\frac{m}{s} }{4 s - 0 s}$

the average acceleration for the first time interval is

$\bar{a}=14\frac{m}{s^{2} }$

ii) The average acceleration for the second time interval, from $t_{1} =4\ s$ to $t_{2} =10\ s$ , is zero because the velocity is constant. The average acceleration is the rate of change of velocity, if this magnitude remains constant then it follows that the acceleration is zero.

iii) We calculate the instantaneous acceleration, which is the acceleration at a specific moment in time, as the derivative of the velocity function.

Mathematically $a(t)=\frac{d v(t)}{d t }$

so we get

$a(t)=-10 \frac{m}{s^{2} } + 12\frac{m}{s^{3} } t$

iv) To find the position in function of time x(t) we simply integrate the expression of v(t):

$v(t)=\frac{d x(t)}{d t}= -10\frac{m}{s^{2} }t+ 6\frac{m}{s^{3} } t^{2}$

we get that

$x(t)=-5\frac{m}{s^{2} } t^{2} + 2 \frac{m}{s^{3} } t^{2}$

Answer from: Quest

what is the actual question though?

Answer from: Quest

force is based upon mass and acceleration.the statement is not always true.

it can be explained from newton's second law of motion.

as per newton's second law of motion the rate of change of momentum is directly proportional to the applied force and takes place along the direction of force.

mathematically it can be written as -

$f\alpha \frac{dp}{dt}$

⇒ $f= k \frac{dp}{dt}$

here k is the proportionality constant whose value is 1.

$f=\frac{dp}{dt}$

$f= \frac{d[mv]}{dt}$

$f= m\frac{dv}{dt} +v\frac{dm}{dt}$

$f=ma+v\frac{dm}{dt}$ [a is the acceleration=dv/dt]

hence we see that force not only depends on mass and acceleration but also on velocity and rate of change of mass. force f will depend on mass and acceleration only when mass is constant.

hence the statement that force is based upon mass and acceleration is false.

Another question on Physics

Physics, 22.06.2019 05:00

When does your body conduct current more readily?

Answers: 1

Answer

Physics, 22.06.2019 06:00

The frequency of vibrations of a vibrating violin string is given by f = 1 2l t ρ where l is the length of the string, t is its tension, and ρ is its linear density.† (a) find the rate of change of the frequency with respect to the following. (i) the length (when t and ρ are constant) (ii) the tension (when l and ρ are constant) (iii) the linear density (when l and t are constant) (b) the pitch of a note (how high or low the note sounds) is determined by the frequency f. (the higher the frequency, the higher the pitch.) use the signs of the derivatives in part (a) to determine what happens to the pitch of a note for the following. (i) when the effective length of a string is decreased by placing a finger on the string so a shorter portion of the string vibrates df dl 0 and l is ⇒ f is ⇒ (ii) when the tension is increased by turning a tuning peg df dt 0 and t is ⇒ f is ⇒ (iii) when the linear density is increased by switching to another string df dρ 0 and ρ is ⇒ f is ⇒

Answers: 3

Answer

Physics, 22.06.2019 07:40

Which lists the fundamental forces in order, from strongest to weakest? strong nuclear, weak nuclear, electromagnetic, gravitational strong nuclear, electromagnetic, weak nuclear, gravitational gravitational, weak nuclear, electromagnetic, strong nuclear electromagnetic, gravitational, strong nuclear, weak nuclear

Answers: 2

Answer

Physics, 22.06.2019 08:30

If an astronaut takes an object to the moon the following is true (choose only one). a the mass will change but the weight will be the same. b both the mass and the weight of the object will be the same. c the weight will change but the mass will be the same. d both the mass and the weight of the object will change.

Answers: 1

Answer

You know the right answer?

Starting from rest at the origin, x0 = 0, a car accelerates in a straight line along the +x directio...

Questions