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b.2 Equations used

v

^{2}=u

^{2}+2as

s=ut+1/2at

^{2}

and other equations

## The Attempt at a Solution

distance = v

^{2}=u

^{2}+2as

s= v

^{2}- u

^{2}/2a

s= 263.1m

The answer is t = 11.2s

....

I am really stuck guys. I understand the final velocity of the entering car must be greater to catch up to the car travelling constant acceleration, but the displacement is the same. But I am finding it difficult to proceed without knowing the final velocity of entering car or the displacement.

Can someone offer a worked example for me - with different values?