can anyone can answer the 10 11 12
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10)--
We have,
distance (s) = ?
Time = (t) = 4 seconds
Acceleration (a) = 2.5m/s²
Initial velocity (u) = 0 (the girl started the race)
So, by using second equation of motion we have,
s = ut + at²
s = 0*4 + 2.5*4*4
s = 0 + 20
s = 20
So, the distance travelled by the girl is 20 m.
11)
ANS-
Given
a = -2m/s2
u = 40m/s
s = 400m
v = 0 (as train stops) by applying s = ut + 1/2 at2
400 = 40t - 1/2*2t2
t = 20 sec
12)
ANS-
a. What is a cheetah’s average acceleration between rest and 20 m s–1?
u = 0, v = 20 ms-1, t = 2 sec, a = ?
v = u + at
a = (v-u)/t = (20-0) / 2 = 10 ms – 2.
b. Assume that a cheetah accelerates up to its top speed with the acceleration in your answer to a.
(i)How far will the cheetah travel when it accelerates from rest up to its top speed?
a = 10 ms – 2, u = 0, v = 30 ms-1, S = ?
v2 = u2 + 2aS
S = (v2 - u2) / 2a = (302 - 02) / (2x10) = 45 m.
(ii) How long does this acceleration take?
v = u + at
t = ( v – u ) / a = ( 30 – 0 ) / 10 = 3 sec.
c. If the cheetah continues at top speed, how long will it be ( from the start of chase ) before it has to stop to rest?
At top speed, u = 30 ms-1, S = 450 m, t = ?
u = S / t
t = S / u = 450 / 30 = 15 sec.
Therefore, Total Time of chase before it has to stop = (Cheetah’s accelerating time) + (Cheetah’s time for uniform motion at top speed) = 3 + 15 = 18 sec.
d. If an antelope starts from rest and accelerates to its top speed at the same rate as a cheetah, how far will it travel in the time obtained in your answer to c?
When antelope accelerates from rest to top speed at same rate as the cheetah,
u = 0, v = 22 ms-1, a = 10 ms – 2, t = ? , S = ?
v = u + at
t = ( v – u ) / a = ( 22 – 0 ) / 10 = 2.2 sec.
Therefore, S = ut + ½. at 2 = 0 + ½. 10 . (2.2)2 = 24.2 m.
Therefore, remaining time = 18 – 2.2 = 15.8 sec.
When antelope runs in the remaining time at top speed of 22ms-1,
u = 22 ms-1, t = 15.8 sec , S = ?
u = S / t
S = ut = 22 x 15.8 = 347.6 m
Therefore the total distance covered by the antelope in 18 sec = 24.2 + 347.6 = 371.8 m.
e. If a cheetah chases an antelope and both start from rest, what is the maximum head start the cheetah can allow the antelope?
the total distance covered by the antelope in 18 sec = 371.8 m.
the total distance covered by the cheetah can in 18 sec = 45 + 450 = 495 m.
Therefore, the maximum head start the cheetah can allow the antelope = 495 – 371.8
= 123.2 m.
hope it helps u!!
We have,
distance (s) = ?
Time = (t) = 4 seconds
Acceleration (a) = 2.5m/s²
Initial velocity (u) = 0 (the girl started the race)
So, by using second equation of motion we have,
s = ut + at²
s = 0*4 + 2.5*4*4
s = 0 + 20
s = 20
So, the distance travelled by the girl is 20 m.
11)
ANS-
Given
a = -2m/s2
u = 40m/s
s = 400m
v = 0 (as train stops) by applying s = ut + 1/2 at2
400 = 40t - 1/2*2t2
t = 20 sec
12)
ANS-
a. What is a cheetah’s average acceleration between rest and 20 m s–1?
u = 0, v = 20 ms-1, t = 2 sec, a = ?
v = u + at
a = (v-u)/t = (20-0) / 2 = 10 ms – 2.
b. Assume that a cheetah accelerates up to its top speed with the acceleration in your answer to a.
(i)How far will the cheetah travel when it accelerates from rest up to its top speed?
a = 10 ms – 2, u = 0, v = 30 ms-1, S = ?
v2 = u2 + 2aS
S = (v2 - u2) / 2a = (302 - 02) / (2x10) = 45 m.
(ii) How long does this acceleration take?
v = u + at
t = ( v – u ) / a = ( 30 – 0 ) / 10 = 3 sec.
c. If the cheetah continues at top speed, how long will it be ( from the start of chase ) before it has to stop to rest?
At top speed, u = 30 ms-1, S = 450 m, t = ?
u = S / t
t = S / u = 450 / 30 = 15 sec.
Therefore, Total Time of chase before it has to stop = (Cheetah’s accelerating time) + (Cheetah’s time for uniform motion at top speed) = 3 + 15 = 18 sec.
d. If an antelope starts from rest and accelerates to its top speed at the same rate as a cheetah, how far will it travel in the time obtained in your answer to c?
When antelope accelerates from rest to top speed at same rate as the cheetah,
u = 0, v = 22 ms-1, a = 10 ms – 2, t = ? , S = ?
v = u + at
t = ( v – u ) / a = ( 22 – 0 ) / 10 = 2.2 sec.
Therefore, S = ut + ½. at 2 = 0 + ½. 10 . (2.2)2 = 24.2 m.
Therefore, remaining time = 18 – 2.2 = 15.8 sec.
When antelope runs in the remaining time at top speed of 22ms-1,
u = 22 ms-1, t = 15.8 sec , S = ?
u = S / t
S = ut = 22 x 15.8 = 347.6 m
Therefore the total distance covered by the antelope in 18 sec = 24.2 + 347.6 = 371.8 m.
e. If a cheetah chases an antelope and both start from rest, what is the maximum head start the cheetah can allow the antelope?
the total distance covered by the antelope in 18 sec = 371.8 m.
the total distance covered by the cheetah can in 18 sec = 45 + 450 = 495 m.
Therefore, the maximum head start the cheetah can allow the antelope = 495 – 371.8
= 123.2 m.
hope it helps u!!
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