Question #76c41

Your goal here is to find the acceleration of the car so that you can use the equation

#color(blue)(ul(color(black)(d = v_0 * t + 1/2 * a * t^2)))#

Here

• #d# is the distance traveled in the #"3 s"#
• #v_0# is the initial velocity of the car
• #t# is the time given
• #a# is the acceleration of the car

Now, you know that an object's acceleration tells you how said object's velocity is changing with respect to time.

#a = (Deltav)/(Deltat) color(white)(color(blue)( larr " change in velocity")/(color(purple)(larr" change in time"))#

In your case, you know that the velocity of the car changes by

#Deltav = "25 m s"^(-1) - "15 m s"^(-1)#

#Deltav = "10 m s"^(-1)#

and that it takes #"3 s"# for this change to occur. If you take #t=0# to be the time when the car begins to accelerate, you can say that the change in time is equal to

#Deltat = "3 s"- "0 s"#

#Deltat = "3 s"#

This means that the car's acceleration is equal to

#a = "10 m s"^(-1)/"3 s" = 10/3 color(white)(.)"m s"^(-2)#

This means that with every passing second, the velocity of the car increases by #10/3color(white)(.)"m s"^(-1)#.

Plug this into the first equation and calculate #d#

#d = "15 m" color(red)(cancel(color(black)("s"^(-1)))) * 3color(red)(cancel(color(black)("s"))) + 1/2 * 10/3color(white)(.)"m" color(red)(cancel(color(black)("s"^(-2)))) * 3^2 color(red)(cancel(color(black)("s"^(2))))#

#d = "45 m" + "45 m" = color(darkgreen)(ul(color(black)("90 m")))#

Therefore, you can say that the car travels #"90 m"# in the #"3 s"# it's being accelerated.