An example would be 2 types of motion. It could be rectilinear or projectile motion. There are various equations for each type. Since you don't want me to tell you the answer, I could just express it in words. Then, it will be up to you to translate into mathematical equations.

For rectilinear motion, the distance traveled is equal to the initial velocity times the time, plus one-half of the acceleration times the square of the time. For projectile motion, the maximum distance is equal to the square of the initial velocity multiplied with the square of the sine of the launch angle, all over twice the gravity.

5 0

3 years ago

Hey there!

eimsori [14]

The time taken for the two balls to hit each other is 8 s.

The given parameters:

Acceleration of the rocket, a = 2 m/s²
Length of the chamber, s = 4 m
Speed of the first ball, = V1 = 0.3 m/s
Speed of the second ball, V2 = 0.2 m/s

The time taken for the two balls to hit each other is calculated by applying relative velocity formula as shown below;

$(V_1 - (-V_2) )t = s\\\\(V_1 + V_2) t = s\\\\(0.3 + 0.2) t = 4\\\\0.5t = 4\\\\t = \frac{4}{0.5} \\\\t = 8 \ s$

Thus, the time taken for the two balls to hit each other is 8 s.

Learn more about relative velocity here: brainly.com/question/17228388

8 0

2 years ago

Read 2 more answers

Suppose that the space shuttle Columbia accelerates at 14.0 m/s2 for 8.50 minutes after takeoff.

givi [52]

Answer:

A. speed = 7.14 Km/s

B. distance = 1820.7 Km

Explanation:

Given that: a = 14.0 m/ $s^{2}$ , t = 8.50 minutes.

But,

t = 8.50 = 8.50 x 60

= 510 seconds

A. By applying the first equation of motion, the speed of the shuttle at the end of 8.50 minutes can be determined by;

v = u + at

where: v is the final velocity, u is the initial velocity, a is the acceleration and t is the time.

u = 0

So that,

v = 14 x 510

= 7140 m/s

The speed of the shuttle at the end of 8.50 minute is 7.14 Km/s.

B. the distance traveled can be determined by applying second equation of motion.

s = ut + $\frac{1}{2}$ a $t^{2}$

where: s is the distance, u is the initial velocity, a is the acceleration and t is the time.

u = 0

s = $\frac{1}{2}$ a $t^{2}$

= $\frac{1}{2}$ x 14 x $(510)^{2}$

= 7 x 260100

= 1820700 m

The distance that the shuttle has traveled during the given time is 1820.7 Km.

5 0

2 years ago

A playground slide is in the form of an arc of a circle with a maximum height of 3.0 m, with a radius of 8.5 m, and with the gro

julia-pushkina [17]

Answer:

a) $s \approx 6.676\,m$

Explanation:

a) Let assume that the ground is not inclined, since the bottom of the playground slide is tangent to ground. Then, the length of given by the definition of a circular arc:

$s = \frac{\pi}{4}\cdot R$

$s=\frac{\pi}{4}\cdot (8.5\,m)$

$s \approx 6.676\,m$

The bottom of the slide has a height of zero. The physical phenomenon around Dr. Ritchey's daughter is modelled after Principle of Energy Conservation. The child begins at rest:

$U_{g,A} = K_{B} + W_{fr}$

$m\cdot g \cdot h_{A} = \frac{1}{2}\cdot m \cdot v_{B}^{2} + f\cdot s$

The average frictional force is cleared within the expression:

$f = \frac{m\cdot (g\cdot h_{A}-\frac{1}{2}\cdot v_{B}^{2})}{s}$

$f = \frac{(12\,kg)\cdot [(9.807\,\frac{m}{s^{2}} )\cdot (3\,m)-\frac{1}{2}\cdot (4.5\,\frac{m}{s} )^{2} ]}{6.676\,m}$

$f = 34.684\,N$

4 0

3 years ago