Question

A record of travel along a straight path is as follows:
1. Start from rest with a constant acceleration of
2.77 m/s^2 for 15.0 s.
2. Maintain a constant velocity for the next 2.05 min.
3. Apply a constant negative acceleration of 29.47 m/s^2
for 4.39 s.
(a) What was the total displacement for the trip?
(b) What were the average speeds for legs 1, 2, and 3 of
the trip, as well as for the complete trip?

Answer 1

For first 15sec
Given u=0
a=2.77m/s
t=15sec
S1=?
We know
S1=ut+(1/2)at^2
S1=?

Answer 2

Response plz!!!

Answer 3

I wanna do 2,3 for u

Answer 4

1/2at^2?

Answer 5

This is (1/2)*a*t^2=0.5*a*t^2
Here a=acceleration
t=time

Answer 6

Anyway dont u know this equation
s=ut+(1/2)at^2

Answer 7

Can i start 2 number question?

Answer 8

ok

Answer 9

S1=ut+(1/2)at^2
i've got 312 m/s. is it correct?

Answer 10

I think u r correct

Answer 11

2.
Given t=2.05*60sec
Required equation is
v=u+at=0+2.77*15=?
Now
S=vt=2.77*15*2.05*60m=?

Answer 12

=5110.65?

*can't open my account -_-

Answer 13

Did u get the above solution?

Answer 14

3.
S=ut-.5at^2
S=2.77*15*4.39-.5*29.47*4.39*4.39
S=?

Answer 15

Now add s of 1,2and3

Answer 16

a)
U will get total distance travelled

Answer 17

Response plz