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3 years ago

If an object starting from rest falls 5.7 seconds how far does it fall

1 answer:

5 0

Kinematic question!
Start with your knowns!

$t = 5.7s \\ v0 = 0 \frac{m}{s} \\ a = - 9.8 \frac{m}{ {s}^{2} }$
we want to find x!

the only kineamtic that works with this info is the third!
$x = v0 \times t + \frac{1}{2} \times a \times {t}^{2}$
so we plug in our values and get
$x = 0 + \frac{1}{2} \times ( - 9.8 \frac{m}{ {s}^{2} } ) \times ( {(5.7s)}^{2} )$
this'll equal out to -159.2 meters.

our units equal out, so we know our answer is correct!

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A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle

MAVERICK [17]

Answer:

t₂ = 3.89 s

Explanation:

given,

speed of car = 23 m/s

speed of motorcycle = 23 m/s

after time of 4 s distance between them is equal to = 53 m

motorcycle accelerates at = 7 m/s

time taken to catch up with car = ?

let t₂ be the time in which motorcycle catches car.

distance traveled by car in t₂ s

d = 23 t₂ + 53

distance traveled by motorcycle

using equation of motion

$s = ut + \dfrac{1}{2}at^2$

$s = 23 t_2 + \dfrac{1}{2}\times 7 \times t_2^2$

now, equating both the distances

$23t_2 + 53= 23 t_2 + \dfrac{1}{2}\times 7 \times t_2^2$

$t_2^2 = 15.143$

t₂ = 3.89 s

time taken by the motorcycle to catch the car is equal to 3.89 s

4 0

3 years ago

A 95kg fullback (football player for those not into sports) moving south with a speed of 5.0 m/s has a perfectly inelastic colli

Lunna [17]

Answer:

a. $v=3.11mls$ , $29.4^{0}$

b. $K.E =-697.8J$

Explanation:

To calculate the values in the question, a deep understanding of perfect inelastic collision is important.

When two bodies undergo inelastic collision, two important parameters must be well understood i.e

Momentum: the momentum is always conserved in perfectly inelastic collision. i.e the total momentum after collision is the sum of the individual momentum before collision

Kinetic energy: Kinetic energy is not conserved due to dissipative force.

a.To calculate the velocity, we first find the total momentum before collision

Momentum of player 1 $p_{1} =mv=95kg*5m/s\\p_{1} =475kgm/s\\$

Momentum of player 2 $p_{2} =mv=90kg*3m/s\\p_{1} =270kgm/s\\$

Hence the total momentum $p_{12}=p_{1}+p_{2}\\$

Note, since the direction of movement before collision is due south and due north respectively we have to represent the velocity using the rectangular coordinate

Hence $p_{12}=(m_{1}+m_{2})v=p_{1}i+p_{2}j\\$

$(95+90)v=475i+270j\\$

$v=2.57i+1.45j\\$

solving for the resultant velocity, we have

$v=\sqrt{2.75^{2} +1.45^{2}}\\ v=3.11mls$

To calculate the direction of movement, we have

$\alpha =tan^{-1}=\frac{v_{j} }{v_{i}}\\ \alpha =tan^{-1}=\frac{1.45}{2.57}\\\alpha =29.4^{0}$

b. to calculate the decrease in total kinetic energy, before collision, the total kinetic was

$K.E_{initial} =\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}.\\K.E_{initial} =((1/2)*95*5^{2})+((1/2)*90*3^{2})\\K.E_{initial} =1187.5+405\\K.E_{initial} =1592.5J\\$