Problem (a) A race car starting from rest accelerates at a constant rate of 5.00

Question

Problem (a) A race car starting from rest accelerates at a constant rate of 5.00 m/s2. What is the velocity of the car after it has traveled 1.00 102 ft? (b) How much time has elapsed? (c) Calculate the average velocity two different ways. Strategy (a) We've read the problem, drawn the diagram in the figure, and chosen a coordinate system (steps 1 and 2). We'd like to find the velocity v after a certain known displacement ?x. The acceleration a is also known, as is the initial velocity v0 (step 3, labeling, is complete), so the third equation in this table looks most useful for solving part (a). Given the velocity, the first equation in the table can then be used to find the time in part (b). Part (c) requires substitution into the following equations. (1) v ? ?x ?y = xf ? xi tf ? ti (2) v = v0 + v 2 SOLUTION (a) Convert units of ?x to SI. 1.00 102 ft = (1.00 102 ft) 1 m 3.28 ft = 30.5 m Write the kinematics equation for v2 (step 4). v2 = v02 + 2a?x Solve for v, taking the positive square root because the car moves to the right (step 5). v = v02 + 2a?x Substitute v0 = 0, a = 5.00 m/s2, and ?x = 30.5 m. v = v02 + 2a?x = 0 2 + 2 5.00 m/s2 30.5 m = 17.5 m/s (b) How much time has elapsed? Apply the equation to the right. v = at + v0 Substitute values and solve for time t. 17.5 m/s = 5.00 m/s2 t t = 17.5 m/s 5.00 m/s2 = 3.50 s (c) Calculate the average velocity in two different ways. Apply the definition of average velocity, Equation (1) above. v = xf ? xi tf ? ti = 30.5 m 3.50 s = 8.71 m/s Apply the definition of average velocity, Equation (2) above. v = v0 + v 2 = 0 + 17.5 m/s 2 = 8.75 m/s The answers are easy to check. An alternate technique is to use ?x = v0t + 1/2 at2 to find t and then use the equation v = v0 + at to find v. Notice that the two different equations for calculating the average velocity, due to rounding, give slightly different answers. Question: What is the final speed if the displacement is increased by a factor of 4?

Explanation / Answer

question 4:

1 ft = 0.3048 m

Initial Displacemet = 1.00 102 ft = 30.48 m

Displacemet is increased by factor 4 = 30.48*4 =121.92 m

Thus :

V^2-u^2=2*a*s

V^2 = 2*5*121.92

V=sqrt(2*5*121.92)

V = 34.92 m/s OR 35 m/s

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