Answer to Question #181613 in Classical Mechanics for kan

Question #181613

a car accelerate from rest to reach a certain speed in 10 min. it then continues at this speed for another 10 minutes and decelerate to rest in another 5 min, the total distance covered is 17.5km. find the steady speed reached


1
Expert's answer
2021-04-18T19:32:43-0400

Let the times of motion during three different parts of the distance be t1,t2,t3t_1, t_2, t_3, the constant speed after the acceleration v1v_1, and accelerations during first and third intervals be a1,a3a_1, a_3. Then the corresponding displacements and speeds during three parts of the way are:

1) s1=a1t122s_1 = \frac{a_1 t_1^2}{2}, v1=a1t1v_1 = a_1 t_1

2) s2=v1t2s_2 = v_1 t_2

3) s3=v1t3a3t322,0=v1a3t3s_3 = v_1 t_3 - \frac{a_3 t_3^2}{2}, 0 = v_1 - a_3 t_3.

Total distance is s=s1+s2+s3=a1t122+v1(t2+t3)a3t322s = s_1 + s_2 + s_3 = \frac{a_1 t_1^2}{2} + v_1 (t_2 + t_3) - \frac{a_3 t_3^2}{2}.

From 1) and 3), v1=a1t1=a3t3v_1 = a_1 t_1 = a_3 t_3, from where a3=t1t3a1a_3 = \frac{t_1}{t_3} a_1. Substituting the last expression for a3a_3 and v1=a1t1v_1 = a_1 t_1 from 1) into ss, obtain:

s=a1t122+a1t1(t2+t3)a1t1t32s = \frac{a_1 t_1^2}{2} + a_1 t_1(t_2 + t_3) - \frac{a_1 t_1 t_3}{2}, from where a1=st122+t1t2+t1t32=0.1kmmin2a_1 = \frac{s}{\frac{t_1^2}{2} + t_1 t_2 + \frac{t_1 t_3}{2}} = 0.1 \frac{km}{min^2}.

The steady speed is hence v1=a1t1=0.1kmmin210min=1kmminv_1 = a_1 t_1 = 0.1 \frac{km}{min^2} \cdot 10 min = 1\frac{km}{min}.


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