Question

A truck pulls a trailer from a standstill with a constant force. After 2 minutes, the truck is moving at 48 km/hr. If the same truck applies the same constant force to pull a second trailer which is 3 times the mass of the first trailer, how fast will the truck be moving after 2 minutes?
A.
8 km/hr
B.
16 km/hr
C.
24 km/hr
D.
64 km/hr

Answer 1

@SolomonZelman

Answer 2

@ags2658

Answer 3

b i think

Answer 4

basically \[48 \div 8 =16\]

Answer 5

i mean 48 div 3 = 16

Answer 6

The inital velocity for track 1 is assumed to be 0?

Answer 7

so is it b

Answer 8

I am asking a question about a problem, not giving the answer... but in any case,

Answer 9

no thats 48

Answer 10

the force is constant, that means the acceleration is constant
(b/c you couldn't be possibly losing mass)

Answer 11

So, after two min it has accelerated from v=0 km/h, to v=48 km/h.

Answer 12

yes you can its holding 3 times the mass

Answer 13

so it losses spped because he is going at the same spped he did with 1 trailer so now hes holding 3 times the first trailer tat way you have to divide

Answer 14

\(\large\color{black}{ \displaystyle a=\Delta v (m/s)/\Delta t(s)=48/2(m/s^2)=24(m/s^2) }\)

Answer 15

how

Answer 16

The force is same, but the mass is three times bigger, so that means the acceleration is 3 times slower, for track 2.

Answer 17

\(\large\color{black}{ \displaystyle a_{\rm ~track~2}=24(m/s^2) /3=8(m/s^2) }\)

Answer 18

so its c

Answer 19

You can use this equation to figure out the answer:
\[F=ma\]
Let's say the force of a random object is 10 N, the acceleration is 2 m/s/s and the mass is 5 kg.
\[10 N= (5 kg)(2 m/s/s)\]
Let's multiply the mass by three:
5 X 3=15 kg
Plugging that into our equation, we get:
\[(15 kg)(? m/s/s)=10 N\]
Dividing both sides by 15, you get:
0.6666667 m/s/s
The answer to your problem must be smaller...

Answer 20

my units are off, km/h excuse me

Answer 21

ok now im confused

Answer 22

I simply gave you an example. :D

Answer 23

After two minutes, what is the velocity if it accelerates by 8km/h, and initial velocity is 0?

Answer 24

basically.

Answer 25

I wanted to show you WHY the speed of the second object will be slower. :)

Answer 26

so b

Answer 27

Wait, my units might be off somewhere I have to check

Answer 28

oh so a

Answer 29

yes, I am wrong my units are off.

Answer 30

ok now i get the answer is a,b,c, or d

Answer 31

i said that bec im cofused

Answer 32

\(\large\color{black}{ \displaystyle a_{~\rm track~1}=\Delta v (km/hr)/\Delta t(min)=48(km/hr)/2min }\)

A mass 3 times larger, will be accelerating 3 times slower.
(because force is same: \(m_1a_1=m_2a_2\) )

So your acceleration is:
\(\large\color{black}{ \displaystyle a_{~\rm track~2}=\Delta v (km/hr)/\Delta t(min)~~\div 3 \\ =16(km/hr)/2min }\)

Answer 33

HELP ME PLZ IM SO COFUSED AND THIS IS MY LAST QUESTION

Answer 34

SO B

Answer 35

Correct! :D

Answer 36

Now, in two minutes what will you velocity be?
Velocity = Acceleration • Time

\(\large\color{black}{ \displaystyle a_{~\rm track~2}=16 (km/hr)/2(min)~~\times 2min=16(km/hr) }\)

Answer 37

that was my first answer the whole time

Answer 38

But we need to show why.....

Answer 39

i did the math you know

Answer 40

thx guys i made a 91

Answer 41

Right, like SolomonZelman said. He showed the formulas, and I followed him while putting in my thoughts. If I had made more details, you could have been even MORE confused.

Answer 42

Do you know which ones were wrong? I would like to know if I did something incorrectly. :D

Answer 43

alright guys BYE AND THX FOR THE HELP @ags2658 AND @solomanzelman

Answer 44

@solomonzelman

Answer 45

BYE

Answer 46

You are welcome! Bye! :D