# Mechanics constant acceleration help

Watch
Announcements
#1
I'm stuck on the following question:

A car travels a distance of 100 metres along a straight line in 20 seconds. The car was initially at rest and travelled with uniform acceleration.
A) Find the acceleration of the car.
B) Find the speed of the car when it has travelled 100 metres.

I have the distance 100 metres, time 20 seconds, initial velocity 0ms^-1 but i dont know which formula to put them in in order to get the acceleration of the car. Please help me I don't have long to do it :/ thanks <3
0
5 years ago
#2
(Original post by Sayless)
I'm stuck on the following question:

A car travels a distance of 100 metres along a straight line in 20 seconds. The car was initially at rest and travelled with uniform acceleration.
A) Find the acceleration of the car.
B) Find the speed of the car when it has travelled 100 metres.

I have the distance 100 metres, time 20 seconds, initial velocity 0ms^-1 but i dont know which formula to put them in in order to get the acceleration of the car. Please help me I don't have long to do it :/ thanks <3
You have
s
u
t
and need
a

There is only one SUVAT that uses these
0
#3
(Original post by TenOfThem)
You have
s
u
t
and need
a

There is only one SUVAT that uses these
I got 0.25ms^-2 for a) and 5ms^-1 for b) is that a correct calculation?
0
5 years ago
#4
(Original post by Sayless)
I got 0.25ms^-2 for a) and 5ms^-1 for b) is that a correct calculation?
Not what I got

Can you show working
1
#5
(Original post by TenOfThem)
Not what I got

Can you show working
s=100
t=20
u-0
velocity = 100/20 = 5ms^-1
5-0/20 = 0.25ms^-2
0
5 years ago
#6
(Original post by Sayless)
s=100
t=20
u-0
velocity = 100/20 = 5ms^-1
5-0/20 = 0.25ms^-2

What formulae did you use

It looks as though you have said you cannot do that

Find a first
0
#7
(Original post by TenOfThem)
What formulae did you use

It looks as though you have said you cannot do that

Find a first
s=ut +1/2at^2
s=0x20 +1/2a x 20^2
100=1/2a x 400
1/4=1/2a
0.50ms^2 = a

is that correct?
0
5 years ago
#8
(Original post by Sayless)
s=ut +1/2at^2
s=0x20 +1/2a x 20^2
100=1/2a x 400
1/4=1/2a
0.50ms^2 = a

is that correct?
yes

Now use v=u+at to get v
0
#9
(Original post by TenOfThem)
yes

Now use v=u+at to get v
I got 0+0.50x20 = 10ms^-1 so speed is 0.50 ms^-1 is that right?
0
5 years ago
#10
(Original post by Sayless)
I got 0+0.50x20 = 10ms^-1 so speed is 0.50 ms^-1 is that right?
No

Why have you turned your 10 into 0.5?
0
#11
(Original post by TenOfThem)
No

Why have you turned your 10 into 0.5?
meant to say speed is 10ms-1, the acceleration is 0.5
0
5 years ago
#12
(Original post by Sayless)
meant to say speed is 10ms-1, the acceleration is 0.5
yes
0
X

new posts
Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### Current uni students - are you thinking of dropping out of university?

Yes, I'm seriously considering dropping out (99)
13.56%
I'm not sure (32)
4.38%
No, I'm going to stick it out for now (230)
31.51%
I have already dropped out (17)
2.33%
I'm not a current university student (352)
48.22%