Answer to Question #101286 in Mechanics | Relativity for Rohan

Question #101286
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?
1
Expert's answer
2020-01-20T05:24:48-0500

Solution.

(а) According to the condition of the problem


mv22=Pt\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=2Ptm=100×103t1.2×103=10t1.2v=\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=dvdt=51.2ta=\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


maμN    maμmg    aμgma\le \mu N \implies ma\le \mu mg \implies a \le \mu g

51.2t0.4×10    11.2t0.8    1.2t1.25\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

1.2t1.5625    t1.56251.2    t1.31.2t \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=1.3st=1.3s

Answer. (a)


v=10t1.2v=10\sqrt{\frac{t}{1.2}}

(b)


a=dvdt=51.2ta=\frac{dv}{dt}=\frac{5}{\sqrt{1.2t}}

(c)


t=1.3st=1.3s


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