(a) [tex]-3.5 m/s^2[/tex]
The car's acceleration is given by
[tex]a=\frac{v-u}{t}[/tex]
where
v = 0 is the final velocity
u = 14 m/s is the initial velocity
t = 4 s is the time elapsed
Substituting,
[tex]a=\frac{0-14}{4}=-3.5 m/s^2[/tex]
where the negative sign means the car is slowing down.
(b) 5.7 s
We can use again the same equation
[tex]a=\frac{v-u}{t}[/tex]
where in this case we have
[tex]a=-3.5 m/s^2[/tex] is again the acceleration of the car
v = 0 is the final velocity
u = 20 m/s is the initial velocity
Re-arranging the equation and solving for t, we find the time the car takes to come to a stop:
[tex]t=\frac{v-u}{a}=\frac{0-20}{-3.5}=5.7 s[/tex]
(c) [tex]2.9 s[/tex]
As before, we can use the equation
[tex]a=\frac{v-u}{t}[/tex]
Here we have
[tex]a=-3.5 m/s^2[/tex] is again the acceleration of the car
v = 10 is the final velocity
u = 20 m/s is the initial velocity
Re-arranging the equation and solving for t, we find
[tex]t=\frac{v-u}{a}=\frac{10-20}{-3.5}=2.9 s[/tex]
(1) The acceleration of the car will be [tex]a=-3.5\frac{m}{s^2}[/tex]
(2) The time taken [tex]t=5.7s[/tex]
(3) The time is taken by the car to slow down from 20m/s to 10m/s [tex]t=2.9s[/tex]
(1) The acceleration of the car will be calculated as
[tex]a=\dfrac{v-u}{t}[/tex]
Here
u= 14 [tex]\frac{m}{s}[/tex]
[tex]a=\dfrac{0-14}{4} =-3.5\dfrac{m}{s^2}[/tex]
(2) The time is taken for the same acceleration to 20[tex]\frac{m}{s}[/tex]
[tex]a=\dfrac{v-u}{t}[/tex]
[tex]t=\dfrac{v-u}{a}[/tex]
u=20[tex]\frac{m}{s}[/tex]
[tex]t=\dfrac{0-20}{-3.5} =5.7s[/tex]
(3) The time is taken to slow down from 20m/s to 10m/s with the same acceleration
From same formula
[tex]t=\dfrac{v-u}{a}[/tex]
v=10[tex]\frac{m}{s}[/tex]
u=20[tex]\frac{m}{s}[/tex]
[tex]t=\dfrac{10-20}{-3.5} =2.9s[/tex]
Thus
(1) The acceleration of the car will be [tex]a=-3.5\frac{m}{s^2}[/tex]
(2) The time taken [tex]t=5.7s[/tex]
(3) The time is taken by the car to slow down from 20m/s to 10m/s [tex]t=2.9s[/tex]
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