Answer
The difference in the speed acquired by the bare head and the speed acquired by the helmeted head is $~~0.56~m/s$
Work Step by Step
To find the speed that the bare head acquires, we can find the area under the acceleration versus time curve for the bare head.
We can divide this area into six parts, including a triangle ($0~s$ to $0.002~s$), a rectangle ($0.002~s$ to $0.004~s$), a triangle ($0.002~s$ to $0.004~s$), a rectangle ($0.004~s$ to $0.006~s$), a triangle ($0.004~s$ to $0.006~s$), and a triangle ($0.006~s$ to $0.007~s$)
We can calculate each area separately:
$A_1 = \frac{1}{2}(0.002~s)(120~m/s^2) = 0.12~m/s$
$A_2 = (0.002~s)(120~m/s^2) = 0.24~m/s$
$A_3 = \frac{1}{2}(0.002~s)(20~m/s^2) = 0.02~m/s$
$A_4 = (0.002~s)(140~m/s^2) = 0.28~m/s$
$A_5 = \frac{1}{2}(0.002~s)(60~m/s^2) = 0.06~m/s$
$A_6 = \frac{1}{2}(0.001~s)(200~m/s^2) = 0.10~m/s$
We can find the total area for $t=0~ms$ to $t=7~ms$:
$A = (0.12~m/s)+(0.24~m/s)+(0.02~m/s)+(0.28~m/s)+(0.06~m/s)+(0.10~m/s)$
$A = 0.82~m/s$
The speed that the bare head acquires is $~~0.82~m/s$
To find the speed that the helmeted head acquires, we can find the area under the acceleration versus time curve for the helmeted head.
We can divide this area into four parts, including a triangle ($0~s$ to $0.003~s$), a rectangle ($0.003~s$ to $0.006~s$), a triangle ($0.004~s$ to $0.006~s$), and a triangle ($0.006~s$ to $0.007~s$)
We can calculate each area separately:
$A_1 = \frac{1}{2}(0.003~s)(40~m/s^2) = 0.06~m/s$
$A_2 = (0.003~s)(40~m/s^2) = 0.12~m/s$
$A_3 = \frac{1}{2}(0.002~s)(40~m/s^2) = 0.04~m/s$
$A_4 =\frac{1}{2}(0.001~s)(80~m/s^2) = 0.04~m/s$
We can find the total area for $t=0~ms$ to $t=7~ms$:
$A = (0.06~m/s)+(0.12~m/s)+(0.04~m/s)+(0.04~m/s)$
$A = 0.26~m/s$
The speed that the helmeted head acquires is $~~0.26~m/s$
We can find the difference in speeds:
$0.82~m/s-0.26~m/s = 0.56~m/s$
The difference in the speed acquired by the bare head and the speed acquired by the helmeted head is $~~0.56~m/s$
