Question

A car which is accelerating uniformly travels 22.5m in the eighth second of its motion.
a) what is the acceleration of the car?
b)how far does the car travel in 6.0s?

Answer 1

a(t) = k

\[\int_{7}^{8}kdt=22.5\]
\[8k-7k=22.5\]
\[k=22.5\]seems about right to me

Answer 2

but them im torn between that and thinking that the 22.5m is a displacement due to velocity

Answer 3

the answers are a) 3.0 ms^2
b) 54m
i just don't know how to get the answers

Answer 4

v(t) = kt

k(64-49)/2 = 22.5

k(15) = 45

k = 45/15 = 9/3 = 3

Answer 5

then my second idea is better than my first :)

Answer 6

i know that integrating velocity gives me displacement so and that a constant (uniform) acceleration integrates into v(t) = kt if we assume v(0) = 0

Answer 7

essentially:\[a(t)=k\]\[\int k\ dt=kt=v(t)\]
\[\int^{8}_{7}kt\ dt=\frac{k}{2}8^2-\frac{k}{2}7^2=22.5;\ k=3\]

Answer 8

displacement after 6 secs is then: 3(6)^2/2 = 3(18) = 54

Answer 9

thats the math, I got no idea what the physics notation would amount to since that tends to invoke formulas that are derived from the mathing itself

Answer 10

\[v=v _{0}+at\]
The average speed during the eighth second is 22.5 m/s. This is the speed at \[\frac{7 +8}{2}=7.5secs\]
Speed at the end of the eighth second \[22.5\times \frac{8}{7.5}=24m/s\]
\[v _{0}=0\]
\[at=v\]
\[a=\frac{v}{t}=\frac{24}{3}=3ms ^{-2}\]

Answer 11

The average speed between t = 0 and t = 6 secs is
\[\frac{0+(3\times 6)}{2}=9m/s\]
Distance travelled in 6 secs = average speed * 6 = 9 * 6 = 54 meters

Answer 12

Sorry. There is a typo in the last line of my first posting.\[a=\frac{v}{t}=\frac{24}{8}=3ms ^{-2}\]

Answer 13

@emmykim do you follow?

Answer 14

Why not use this relation...
Distance traveled in nth second is given by S= u +(1/2)(2n-1)a
here u=0
Putting s=22.5 m and n=8
we get a=3m/s^2
using this value of acceleration you will get distance traveled in 6 sec to be 54 m..

As simple as that...

Answer 15

@ujjwal and isn't that the method that I used but with more explanation.