Question

a car traveling at a speed of 13 meters per second accelerates uniformly to a speed of 25 meters per second in 5.0 seconds.
-a truck traveling at a constant speed covers the same total distance as the car in the same 5.0 second interval. Determine the speed of the truck.
-Calculate the magnitude of the accelerate of the car during this 5.0 second interval.

Answer 1

The car:
accelerate \[\large a = \dfrac{v_{final}-v_{initial}}{\triangle t} = \dfrac{25-13}{5}= 6 ~~m/s^2\]
so, the distance it covers in 5 seconds is
\[\large S = \dfrac{v_{final}^2-v_{initial}^2}{2a}=\dfrac{25^2-13^2}{12}=38~~m\]
The truck:
On the same distance with the car 38m, in 5seconds , its speed is
\[\large v= \dfrac{distance}{time}=\dfrac{38}{5}=7.6~~m/s\]

Answer 2

@Loser66 Your processes are correct, but methinks the arithmetic is a bit off ^_^

\[a = \frac{\Delta v}{\Delta t} = \frac{v-v_0}{t}
\\ \
\\ \
\\ \ =\frac{25m/s - 13m/s}{5s}\]
\[a = \frac{12}{5}m/s^2\]

Then to find the distance traveled by the car, I'm gonna use the first equation of motion instead of the 4th (both are valid, of course)
\[ s = v_0 t + \frac{1}{2}at^2
\\ =(13m/s)(5s) + \frac{1}{2}\left(\frac{12}{5}m/s^2\right)(5s)^2
\\ \
\\ = (13m/s)(5s)+\left(\frac{6}{5}m/s^2\right) ( 25s^2)
\\ \
\\ \
\\ =(13m/s)(5s)+ (6m/s)(5s)
\\ \
\\ s= 95m\]

And then the velocity of the truck is given, as stated above by
\[v = \frac{\Delta s}{\Delta t} = \frac{95m}{5s}\]

\[ v = 19m/s\]

Answer 3

thanks for correcting me,