AS Applied Unit 9 Kinematics 2 (variable acceleration)

A car starts from the point A. At time ts after leaving A, the distance of the car from A is s m, where s = 30t-0.4t^2, 0 ⩽t⩽ 25. The car reaches point B when t = 25.
a) Find the distance AB.

b) Show that the car travels with a constant acceleration and state the value of this acceleration.

A runner passes through B when t = 0 with an initial velocity of 2 m s−1 running directly towards A. The runner has a constant acceleration of 0.1 m s−2.

c) Find the distance from A at which the runner and the car pass one another.

What have you done / are stuck with?

ive got a) 500m and b) 0.8ms-2 im stuck on c tho. not sure how to do it

For b) a should be negative.

For c) you could form two suvats and solve as theyre both constant acceleration, or use relative motion if youve come across that. See if you can sketch / form the two suvats (if not) and upload if necessary? One has already been given to you.

can u tell me pls? i cant seem to form two suvats that work

One is given to you, so sketch/formulate the other one starting at B with the appropriate initial velocity and acceleration. Obv you need to match the displacements so they are 500 apart at the start or they sum to 500.

sorry i dont get it

why does sc + sr =500 tho? i dont get that

You want to find when they meet, so their displacements must sum to 500.

If the car had travelled 100 and the runner 400, theyd meet or .... Or you could model the displacement of the runner as 500-sc, so initially at 500 and heading towards A and find the point when theyre the same. There are a few ways to pose the problem which is why a good sketch at the start is important to reason about things like this.

You want to find when they meet, so their displacements must sum to 500.

If the car had travelled 100 and the runner 400, theyd meet or .... Or you could model the displacement of the runner as 500-sc, so initially at 500 and heading towards A and find the point when theyre the same. There are a few ways to pose the problem which is why a good sketch at the start is important to reason about things like this.

is the asnwer 31.2?

A back of the envelope calculation has the initial speed of the car at 30m/s and the runners speed is roughly constant (small acceleration) at 2m/s. So they must meet near B. So the time must be a bit less than 25 and the distance from A must be a bit less than 500.

do you know the answer? ill see if i can back engineer it to understand it better

Just write down the suvat for sr as per the previous post then solve the sr+sc=500 quadratic for t. While you may make a mistake, upload what youve tried pls.

ive got Sc = -0.4t^2 and Sr = 2t+0.05t^2. is that correct

No. Sc is given to you, so where is the 30t? You have the car with zero initial velocity accelerating away from B.

is the runner one correct tho?

Sure, thats assuming youre modelling him as at B at t=0 and moving (positive direction) to the car. So solve when sc+sr=500.