Question

A test driver at Incredible Motors, Inc., is testing a new model car having a speedometer calibrated to read rather than . The following series of speedometer readings was obtained during a test run:

Time () 0 2.0 4.0 6.0 8.0 10 12 14 16
Velocity () 0 0 2.0 5.0 10 15 20 22 22
(a) Compute the average acceleration duringeach 2-s interval. Is the acceleration constant?A. Yes
B. No (b) Is it constant during any part of the test run?A. Yes
B. No

Answer 1

(a) No, the acceleration is not constant

(b) Yes, it is constant between 6 s and 12 s

Step-by-step explanation:

Given;

time of motion; t(s) = 0 2 4 6 8 10 12 14 16

velocity (m/s); v = 0 0 2 5 10 15 20 22 22

Average acceleration for t= (2,0) and v = (0,0)

`a = (v_2-v_1)/(t_2-t_1) \\\\a = (0-0)/(2-0) = 0 \ m/s^2`

Average acceleration for t = (4, 2) and v = (2,0)

`a = (v_2-v_1)/(t_2-t_1) \\\\a = (2-0)/(4-2) = 1.0 \ m/s^2`

Average acceleration for t = (6, 4) and v = (5,2)

`a = (v_2-v_1)/(t_2-t_1) \\\\a = (5-2)/(6-4) = 1.5 \ m/s^2`

Average acceleration for t = (8, 6) and v = (10,5)

`a = (v_2-v_1)/(t_2-t_1) \\\\a = (10-5)/(8-6) = 2.5 \ m/s^2`

Average acceleration for t = (10, 8) and v = (15,10)

`a = (v_2-v_1)/(t_2-t_1) \\\\a = (15-10)/(10-8) = 2.5 \ m/s^2`

Average acceleration for t = (12,10) and v = (20,15)

`a = (v_2-v_1)/(t_2-t_1) \\\\a = (20-15)/(12-10) = 2.5 \ m/s^2`

Average acceleration for t = (14,12) and v = (22,20)

`a = (v_2-v_1)/(t_2-t_1) \\\\a = (22-20)/(14-12) = 1.0 \ m/s^2`

Average acceleration for t = (22, 22) and v = (16,14)

`a = (v_2-v_1)/(t_2-t_1) \\\\a = (22-22)/(16-14) = 0 \ m/s^2`

(a) B. No (The acceleration is not constant)

(b) A. Yes (it is constant between 6 s and 12 s)