Mechanics a level maths help

A car is being driven along a road at 25m/a when the driver suddenly notices that there is a fallen tree blocking the road 65m ahead. The driver immediately applies the brakes giving the car a constant deceleration of 5m/s^2. How far infront of the tree does the car come to rest?

Have you tried the question first?

Perhaps it may be worth starting by underlining some words and numbers given in the question?

Post what you've tried as per forum rules

Yes but I keep getting wrong answer

S- 65
U-25
V-0
A-5
T-
But I don’t understand what to do next

Let’s have a little look at your values of S through to t


Well, the question is asking about how far the car stops in front of the tree. 65 m is how far the tree is from the driver the instant he sees it (as stated in the question), so not necessarily how far the car travels as it decelerates from 25 m/s to 0 m/s. Thus you don’t actually have S, the question threw that in there to trick you.

I presume it’s decelerating. What does that tell you about the acceleration?

Acceleration is negative

Yep. That is correct.

Now look back at the question. How is the deceleration described?

Acceleration is negative

So this is what I’ve but the correct answer is 2.5 m
I got 52.08m

I did v^2= u^2 + 2as
0^2= (25)^2 + 2(-5)s
And I got a to be 52.08
Please help

Rearrange what you’ve got to make s the subject

I think you may have mistyped something into your calculator…

Constant

I have but my calculator keeps saying that…

i did this look…

0= (25)^2 + 2(-5)s
0= 625 -10s
-625 = -10s
62.5 = s

That’s how far the car was travelling as it was stopping. The tree was 65 m in front of the car when it started stopping.

What’s the difference in the displacements

Ohhhh thank you so much! I didn’t realise that’s how you do it lol

I think your biggest problem is that certain details in the question elude you.

If you have a physical copy of the question in front of you, it may be worth highlighting or underlining important information given (ofc don’t do this if it’s a question in a textbook supplied to you by your school).

I think your biggest problem is that certain details in the question elude you.

If you have a physical copy of the question in front of you, it may be worth highlighting or underlining important information given (ofc don’t do this if it’s a question in a textbook supplied to you by your school).

I’m also stuck on this question
A ball is thrown vertically upwards with a speed of 29m/s it hits the ground 6 seconds later by modelling ball as particle find the height above the ground from which it was thrown


the answer is 2.4m but I got something else

my working:
s- (working this out)
u- 29
v-0
a- 9.8
t-6
I did this
S= (v+u/ 2) t
s= 29+0/2 (6)
and got 87

The acceleration should be negative again. Think about why, perhaps?

The time given is how long it takes to get to maximum height then hit the floor again. It’s not actually useful to use t = 3 or 6 in this question, because we don’t know how high it was when it was projected, so assuming it was at ground level before is probably wrong.

So what can you now do?

I did this but got -2.4

s-(working out)
u-29
v-
a- -9.8
t- 6

s= ut+ 1/2 at^2
s= 29(6)+1/2(-9.8)(6)^2
s= -2.4