(a) 8.14 m/s^2
Explanation:
Acceleration of the car is given by:
[tex]a=\frac{v-u}{t}[/tex]
where
u = 0 is the initial velocity
v = 24.0 m/s is the final velocity
t = 2.95 s is the time taken
Substituting numbers into the equation, we find
[tex]a=\frac{24.0 m/s-0}{2.95 s}=8.14 m/s^2[/tex]
(b) 1.23 s
Explanation:
We can use the same formula for the acceleration:
[tex]a=\frac{v-u}{t}[/tex]
where this time, we have
u = 10.0 m/s
v = 20.0 m/s
a = 8.14 m/s^2
Substituting numbers into the equation, we can find the time taken, t:
[tex]t=\frac{v-u}{a}=\frac{20.0 m/s-10.0 m/s}{8.14 m/s^2}=1.23 s[/tex]
c) If the acceleration is constant, yes
Explanation:
Let's take again the equation for the acceleration:
[tex]a=\frac{v-u}{t}[/tex]
We can rewrite it as:
[tex](v-u) = at[/tex]
where the term (v-u) represents the change in velocity. From the formula we see that, if a (acceleration) is constant, then (v-u) is directly proportional to the time t: therefore, if t doubles, the change in velocity (v-u) doubles as well.
