Constant Acceleration Question 2

Question: a car travelling at 30 m s^-1 brakes with a constant deceleration of 6.0 m s^-2 until it stops. Calculate (a) the time taken to stop, (b) the stopping distance and (c) the distance travelled in the first 2.0 s of braking.Step 1: choose a sign convention. Take the direction of motion as positive. The car is slowing down, so its acceleration is negative.u = 30 m s^-1, v = 0 (it stops), a = -6.0 m s^-2(a) Time to stop, using v = u + at:0 = 30 + (-6.0)t, so t = 5.0 s(b) Stopping distance, using v² = u² + 2as:0 = 30² + 2(-6.0)ss = 900 ÷ 12 = 75 m(c) First 2.0 s, using s = ut + ½at²:s = (30 × 2.0) + ½(-6.0)(2.0)² = 60 - 12 = 48 mThe car covers 48 m of its 75 m stopping distance in the first 2.0 s, because it is moving fastest at the start.Common mistakes: using a = +6.0 m s^-2 for a deceleration, which makes the speed increase instead of fall; and forgetting that v = 0 when an object comes to rest.