A car is moving along a straight horizontal road - Edexcel - A-Level Maths Mechanics - Question 4 - 2007 - Paper 1

Using the equation for distance, the total distance covered from points A to B is given by:

$ext{Distance} = (Speed_1 imes Time_1) + (Speed_2 imes Time_2) + (Speed_3 imes Time_3)$

where:

• Speed_1 = 25 m/s, Time_1 = 10 s
• Speed_2 = (25 + V) m/s, Time_2 = 8 s
• Speed_3 = V m/s, Time_3 = 12 s

We can set up the equation:

$25 imes 10 + (25 + V) imes 8 + V imes 12 = 526$

Solving this gives:

$250 + 8 imes V + 200 + 12 imes V = 526$

Thus,

\Rightarrow 20V = 76\ \Rightarrow V = 3.8$$ m/s.

To find the deceleration, we can use the equation of motion:

$v = u + at$

where:

• $v = V$ (the final speed after 8 s)
• $u = 25$ (the initial speed at t = 10 s)
• $t = 8$ s, and $a$ is the deceleration.

Substituting the known values gives:

$V = 25 + 8a$

We already found that $V = 11$, so:

\Rightarrow 8a = 11 - 25\ \Rightarrow 8a = -14\ \Rightarrow a = -1.75 m/s^2$$ Thus, the deceleration is 1.75 m/s².