A car starts from Rest and moves along the X-axis with our constant acceleration 5 m/s for 8 seconds. If it then continues with constant velocity, what distances will the car cover in 12 seconds since it started from the rest?
Answers
Answered by
98
In first 8 seconds,
acceleration(a)=v-u/t
as u =0
a=v/t
5*8=v=40m/s.
in this time distance(d)=u*t+1/2*a*t^2
=)+1/2*5*64=160m.
In next 4 seconds,
d=v*t=40*4=160m.
So total d = 160+160=320m
acceleration(a)=v-u/t
as u =0
a=v/t
5*8=v=40m/s.
in this time distance(d)=u*t+1/2*a*t^2
=)+1/2*5*64=160m.
In next 4 seconds,
d=v*t=40*4=160m.
So total d = 160+160=320m
Answered by
78
For the journey with constant acceleration, the distance travelled = ut+ at²/2
= 5×8²/2 = 160 m
u = 0 m/s, because the car starts from rest.
Now let us calculate the velocity after these 8 seconds.
v= u +at = 5×8 =40m/s
Since 8 seconds are spent with constant acceleration, the rest 4 seconds from the 12 seconds are spent with constant velocity.
Now, the distance covered by the car with this constant velocity after seconds =
= speed × time
= 40 × 4
= 160 m
Hence, total distance covered = 160+160= 320m
I hope you understand the concept well.
If you find it helpful please mark it as brainliest.
All the best!
= 5×8²/2 = 160 m
u = 0 m/s, because the car starts from rest.
Now let us calculate the velocity after these 8 seconds.
v= u +at = 5×8 =40m/s
Since 8 seconds are spent with constant acceleration, the rest 4 seconds from the 12 seconds are spent with constant velocity.
Now, the distance covered by the car with this constant velocity after seconds =
= speed × time
= 40 × 4
= 160 m
Hence, total distance covered = 160+160= 320m
I hope you understand the concept well.
If you find it helpful please mark it as brainliest.
All the best!
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