Acceleration is given by:
[tex]a=\frac{v-u}{t}[/tex]
where
v is the final velocity
u is the initial velocity
t is the time interval
Let's apply the formula to the different parts of the problem:
A) [tex]-20.5 m/s^2[/tex]
Let's convert the quantities into SI units first:
[tex]u = 19300 km/h \cdot \frac{1000 m/km}{3600 s/h} = 5361.1 m/s[/tex]
[tex]v=1600 km/h \cdot \frac{1000 m/km}{3600 s/h} =444.4 m/s[/tex]
t = 4.0 min = 240 s
So the acceleration is
[tex]a=\frac{444.4 m/s-5361.1 m/s}{240 s}=-20.5 m/s^2[/tex]
B) [tex]-3.8 m/s^2[/tex]
As before, let's convert the quantities into SI units first:
[tex]u = 444.4 m/s[/tex]
[tex]v=321 km/h \cdot \frac{1000 m/km}{3600 s/h} =89.2 m/s[/tex]
t = 94 s
So the acceleration is
[tex]a=\frac{89.2 m/s - 444.4 m/s}{94 s}=-3.8 m/s^2[/tex]
C) [tex]-53.0 m/s^2[/tex]
For this part we have to use a different formula:
[tex]v^2 - u^2 = 2ad[/tex]
where we have
v = 0 is the final velocity
u = 89.2 m/s is the initial velocity
a is the acceleration
d = 75 m is the distance covered
Solving for a, we find
[tex]a=\frac{v^2-u^2}{2d}=\frac{0^2-(89.2 m/s)^2}{2(75 m)}=-53.0 m/s^2[/tex]
