1) Acceleration of the car: 2.4 m/s^2
Explanation:
The acceleration of the car is given by:
\(a=\frac{v-u}{t}\)
where
v = 25 m/s is the final velocity of the car
u = 13 m/s is the initial velocity
t = 5.0 s is the time taken
Substituting the numbers into the equation, we find
\(a=\frac{25 m/s-13 m/s}{5.0 s}=2.4 m/s^2\)
2) Speed of the truck: 19 m/s
First of all, we need to find the total distance covered by the car, which is given by the equation:
\(d=u t + \frac{1}{2}at^2=(13 m/s)(5.0 s)+\frac{1}{2}(2.4 m/s^2)(5.0 s)^2=95 m\)
The truck covers the same distance in t=5.0 s, travelling at constant speed; therefore, its speed is given by
\(v=\frac{d}{t}=\frac{95 m}{5.0 s}=19 m/s\)
