Answer to Question #101286 in Mechanics | Relativity for Rohan

Question

3. Acceleration [8]
Assume that a car with mass m =1,200 kg accelerates from rest with a constant power (constant change of mv2
energy with time) P = 50. kW. Then the kinetic energy is given by 2 =Pt .
(a) Find the speed as a function of time. [2]
(b) Differentiate the speed to find the acceleration as a function of time. [2]
(c) Assume that the acceleration is limited by the coefficient of friction between the tires and the wet road,
=0.40 . What is the minimum time for the result in part (b) to be valid?

Answer 1

Solution.

(а) According to the condition of the problem

$$\frac {mv^2} {2}=Pt$$

where m=1200kg is mass of the car; v is the speed of the car; P=50000W is a constant power.

Therefofe

$$v=\sqrt{\frac{2Pt} {m}}=\sqrt{\frac{100\times 10^3 t} {1.2\times 10^3}}=10\sqrt{\frac{t}{1.2}}$$

(b) Differentiate the speed to find the acceleration as a function of time.

$$a=\frac{dv}{dt}=\frac{5}{\sqrt{1.2t}}$$

(c) The force acting on the car must be less than or equal to the friction force

$$ma\le \mu N \implies ma\le \mu mg \implies a \le \mu g$$

$$\frac{5}{\sqrt{1.2t}} \le 0.4\times10 \implies \frac {1} {\sqrt{1.2t}} \le 0.8 \implies {\sqrt{1.2t}} \ge 1.25$$

$$1.2t \ge 1.5625 \implies t \le \frac {1.5625} {1.2} \implies t \ge 1.3$$

Hence the minimum time for the result in part (b) to be valid

$$t=1.3s$$

Answer. (a)

$$v=10\sqrt{\frac{t}{1.2}}$$

(b)

$$a=\frac{dv}{dt}=\frac{5}{\sqrt{1.2t}}$$

(c)

$$t=1.3s$$