Standard +0.3 This is a standard M1 power question requiring two applications of F=ma and P=Fv to form simultaneous equations. The method is routine and well-practiced, though the algebra with two unknowns and the need to carefully apply Newton's second law at two different states elevates it slightly above average difficulty.
A car of mass 860 kg travels along a straight horizontal road. The power provided by the car's engine is \(P\) W and the resistance to the car's motion is \(R\) N. The car passes through one point with speed \(4.5 \text{ m s}^{-1}\) and acceleration \(4 \text{ m s}^{-2}\). The car passes through another point with speed \(22.5 \text{ m s}^{-1}\) and acceleration \(0.3 \text{ m s}^{-2}\). Find the values of \(P\) and \(R\). [6]
Question 3:
3 | P
– R = 860 × 4
4.5
P
– R = 860 × 0.3
22.5
P P
– = 860(4 – 0.3) (cid:1)
4.5 22.5
P = 17900
or
–4.5R + 22.5R =
860(4 × 4.5–0.3 × 22.5) (cid:1)
R = 537.5
R = 537.5 | M1
A1
A1
M1
A1
B1 | 6 | For using DF = P/v and for applying
nd
Newton’s 2 law at one or both points
For eliminating R to find P or for
eliminating P to find R
Accept 538
A car of mass 860 kg travels along a straight horizontal road. The power provided by the car's engine is $P$ W and the resistance to the car's motion is $R$ N. The car passes through one point with speed $4.5 \text{ m s}^{-1}$ and acceleration $4 \text{ m s}^{-2}$. The car passes through another point with speed $22.5 \text{ m s}^{-1}$ and acceleration $0.3 \text{ m s}^{-2}$. Find the values of $P$ and $R$. [6]
\hfill \mbox{\textit{CAIE M1 2015 Q3 [6]}}