# M1 Kinematics: 2 solutions , which is correct?

Watch
Announcements
#1

In this question , the acceleration is worked out to be 0.34 ms^-2 in part a).

The problem is that in part b); to work out the total time taken for the car to travel to A to C; I used the suvat equation s=vt -0/5at^2, where s is the total distance; (100+300=400m), v is the final velocity when the car is at point C (20m/s) and acceleration as previously worked out is 0.34ms^-2. So all I have to do is work out t. However when using this suvat equation, I get two values of t; both are positive but only ONE is the right answer; which is 25.5s. (The other value is something like 94s). When using this equation, how could you possible know which one is the correct one, and thinking about it in terms of the question , how could there even be two possible times for the car travelling from A to C , because the car has a positive acceleration and only passes both points once, then accelerating off into the distance further away?
0
6 years ago
#2
(Original post by scientific222)

In this question , the acceleration is worked out to be 0.34 ms^-2 in part a).

The problem is that in part b); to work out the total time taken for the car to travel to A to C; I used the suvat equation s=vt -0/5at^2, where s is the total distance; (100+300=400m), v is the final velocity when the car is at point C (20m/s) and acceleration as previously worked out is 0.34ms^-2. So all I have to do is work out t. However when using this suvat equation, I get two values of t; both are positive but only ONE is the right answer; which is 25.5s. (The other value is something like 94s). When using this equation, how could you possible know which one is the correct one, and thinking about it in terms of the question , how could there even be two possible times for the car travelling from A to C , because the car has a positive acceleration and only passes both points once, then accelerating off into the distance further away?
I don't know what sort of SUVAT equations you've been using, but I only get one value of t which is approx 25.5 seconds.
0
6 years ago
#3
For your second value of t of 94s, the initial velocity is negative, and the car travels backwards from A, and comes to rest, and then starts moving forward, through A, to C.
Acceleration is constant throughout.
0
#4
(Original post by ghostwalker)
For your second value of t of 94s, the initial velocity is negative, and the car travels backwards from A, and comes to rest, and then starts moving forward, through A, to C.
Acceleration is constant throughout.
How have you deduced that the initial velocity is negative and the car travels backwards from A?
0
6 years ago
#5
(Original post by scientific222)
How have you deduced that the initial velocity is negative and the car travels backwards from A?
For suvat (constant acceleleration) the graph of the displacment is a quadratic. If you recall the shape.

If there are two solutions, one's on the up slope and one on the down. Moving forwards, moving backwards.
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

#### Are you travelling in the Uni student travel window (3-9 Dec) to go home for Christmas?

Yes (73)
27.76%
No - I have already returned home (30)
11.41%
No - I plan on travelling outside these dates (54)
20.53%
No - I'm staying at my term time address over Christmas (28)
10.65%
No - I live at home during term anyway (78)
29.66%