Question 1.7: James practises using the stopwatch facility on his new watc...

i) The diagram shows all the information assuming the acceleration is a ms^{-2} and the velocity at A is u ms^{-1}.

For AB, s = 30 and t = 1.2. You are using u and you want a so you use

s = ut  +  \frac{1}{2}at^{2}

30 = 1.2u  + \frac{1}{2}a × 1.2^{2}

30 = 1.2u + 0.72a                                                 ①

To use the same equation for the part BC you would need the velocity at B and this brings in another unknown. It is much better to go back to the beginning and consider the whole of AC with s = 60 and t = 2.2. Then again using s = ut  + \frac{1}{2}at^{2}

60 = 2.2u  +  \frac{1}{2}a × 2.2^{2}

60 = 2.2u + 2.42a                                                   ②

These two simultaneous equations in two unknowns can be solved more easily if they are simplified. First make the coefficients of u integers.

① × 10 ÷ 12              25 = u + 0.6a                        ③

② × 5                        300 = 11u + 12.1a                  ④

then                      ③ × 11                      275 = 11u + 6.6a                     ⑤

Subtracting gives

25 = 0 + 5.5a

a = 4.545

Now substitute 4.545 for a in  ③ to find

u = 25 – 0.6 × 4.545 = 22.273 .

The acceleration of the car is 4.55 ms^{-2} and the initial speed is 22.3 ms{-1} (correct to 3 s.f.).

For this part, you know that s = 30, v = 22.3 and a = 4.55 and you want t so you use the fifth formula.

s = vt  –  \frac{1}{2}at^{2}

30 = 22.3 × t  –  \frac{1}{2} × 4.55 × t^{2}

⇒                   2.275t² – 22.3t + 30 = 0

Solving this using the quadratic formula gives t = 1.61  and  t = 8.19.

The most sensible answer to this particular problem is 1.61 s.