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Constant Acceleration Equations

A bus is approaching a tunnel. At time t=0 seconds, the driver begins slowing down steadily from a speed of Ums-1 until, at t=15s, he enters the tunnel, travelling at 20ms-1. The driver maintains this speed while he drives though the tunnel. After emerging from the tunnel at t=40s, he accelerates steadily, reaching a speed of Ums-1 at t=70s.

A)calculate the length of the tunnel
B) Given that the total distance travelled by the bus is 1580m, find the value of U.
where u getting these questions from lool. It looks like u really r struggling with suvat questions.
Original post by dj_ad_1
where u getting these questions from lool. It looks like u really r struggling with suvat questions.

ive had a bank of questions that im stuck on.... haven't actually gone though it yet in lesson but im just getting ahead
Original post by JasmineJohal14
ive had a bank of questions that im stuck on.... haven't actually gone though it yet in lesson but im just getting ahead

Yh i thought that ngl. Thats acc mad that ur going ahead of everyone maybe if u snap me em ill help u cos like im 2nd yr innit so it would kinda be revision for me ygm
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MLJ_
a)For the length of the tunnel, 'S' is what you want to find in terms of your suvat equations. Your initial velocity (U) should be 20 since he enters the tunnel at this speed and it's maintained so final velocity will also be 20 whilst he's in the tunnel. He entered the tunnel at 15 seconds and emerged at 40 so by finding the difference (40-15), you know that time here is 25. After that, just simply plug your values into the equation, s=(u+v/2)t to find s. I get 500m



b) You need to find the distance before entering the tunnel and after exiting it. Add this with 500m (distance in tunnel) and equate to 1580. Use the equation s=(u+v/2)t for both equations in terms of U.
(edited 4 years ago)

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