The maker of an automobile advertises that it takes 13 seconds to accelerate from 25 kilometers per hour to 90 kilometers per hour. Assuming constant acceleration, compute the following.

Question

The acceleration in m/s^2

The distance the car travels during the 13 seconds

Answer 1

if we have acceleration of a, then
v = at + c
case1: when t=0, v=25
25 = 0+c ---> v = at + 25
so when t = 13, v = 90
90 = 13a + 25
a = 65/13 = 5 m/s^2 -----> v = 5t + 25

distance = (5/2)t^2 + 25t + k
when t=0 , distance = 0 , so k = 0

distance = (5/2)t^2 + 25t
when t = 13
distance = (5/2)(169) + 25(13) = 747.5 m

also see the first four of the Related Questions below

Answer 2

a = (V2-V1)/(t2-t1)

V2 = 90,000/3600 = 25 m/s
V1 = 25,000/3600 = 6.94

t2-t1 = 13

so a = 1.39 m/s^2
======================
d = V1 t + (1/2)a t^2
= 6.94 *13 + .695*169
= 208 m

Answer 3

watch out, km/hour given

Answer 4

forgot to change the units from km/h to m/s

go with Damon's answers