Question

An object starts from rest with an acceleration of 2.0 m/s2 for 3.0 seconds. It then reduces its acceleration to 1.0 m/s2 for 5.0 additional seconds. The total distance covered is

Answer 1

Using these (SUVAT) equations http://en.wikipedia.org/wiki/Equations_of_motion#Constant_linear_acceleration:_collinear_vectors

We find for the 1st 3 seconds, it travels,
$$
r_{3 sec}=r_0+v_0t+{at^2\over2}\\
r={2\times3^2\over2}=9~meters
$$
Its velocity at the end of 3 seconds is
$$
v=at=3\times3=9~m/s
$$
So after 5 seconds, it has traveled (using first equation above)
$$
r_{5 sec}=9+3(5)+{1(5)^2\over 2}~meters
$$

Answer 2

Thank you ybarrap but in the options, the possible answers are 53 meters, 38 meters, 11 meters, 9 meters or 7.6 meters. None of it match your own answer please I need help. Thank you very much!

Answer 3

52 meters I mean.

Answer 4

for what ybarrap has said using the equation V=U+at the final velocity at the end of 3 second is 6m/s2 but not 9

Answer 5

for the distance s2 s=ut +1/2 at^2
s=6(5) +1/2 (-1)5*5
s=`17.5
adding both we get 26.5

Answer 6

For 3 sec use
S=ut+0.5at^2
S=0*3+0.5*2*3^2=?

Answer 7

Now after 3sec the final velocity will b
v=u+at
v=0+2*3
v=6m/s
Right?

Answer 8

Now for additional 5 sec
S=ut+0.5at^2
S=6*5+0.5*1*5^2
S=?

Answer 9

Now add those 2 s. U will get total distance travelled

Answer 10

Response plz!!!!!!

Answer 11

Hi shamim, the answer for S for the 3 sec is 9 m and the S for the additional 5 sec is 42.5 and after adding 9 and 42.5 I got 51.5 which is approximately 52. And that's part of the option that was given in the question. Thank you very much.