a body travels 200 cm in the first 2s and 220 cm in the next 5s. calculate the velocity at the end of the seventh second from the start.
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hey mate
We have a body initially moving at a certain velocity, which we will call uu. Assuming constant acceleration of aa, this body is clearly slowing down, so a<0a<0 - if the body’s speed was constant, i.e. a=0a=0, the distance covered between time t=2 secondst=2 seconds to t=6 secondst=6 seconds would be 4 m4 m; and, if the speed was increasing, i.e. a>0a>0, the distance covered would have been more than 4 m4 m.
The obvious equation to use is:
s=ut+12at2s=ut+12at2, where ss is the distance travelled (in metres)
Now, let’s insert the information from the question into this equation.
A: 2=2u+1222a=2u+2a⇒u+a=12=2u+1222a=2u+2a⇒u+a=1
B: (2+2.2)=(2+4)u+12(2+4)2a(2+2.2)=(2+4)u+12(2+4)2a
⇒4.2=6u+18a⇒u+3a=0.7⇒4.2=6u+18a⇒u+3a=0.7
Subtracting A from B, we have: 2a=−0.3⇒a=−0.152a=−0.3⇒a=−0.15
Substituting this into A, we thus have: u−0.15=1⇒u=1.15u−0.15=1⇒u=1.15
So, what have we got so far?
The initial velocity of the object is 1.15 ms−11.15 ms−1
The object is accelerating at the rate of −0.15 m−2−0.15 m−2
Now, what do we want to find out?
From the question: What is [the object’s] velocity from the start to the end of the seventh second?
What does this mean? Well, there is more than one interpretation!
What seems (at least to me) the most likely meaning is that we have been asked to find the object’s initial velocity and the velocity at time t=7 secondst=7 seconds; in which case, the question should have said “What are the velocities at the start and after 7 seconds?”
Well, we have already calculated the initial velocity, so we’re half-way there!
The equation we need to determine the velocity, vv, at a certain time is:
v=u+atv=u+at
Feeding in the information we know: v=1.15+(−0.15)×7v=1.15+(−0.15)×7
=1.15−1.05=0.10=1.15−1.05=0.10
Thus the velocity after 7 seconds is 0.1 ms−10.1 ms−1
Answers:
Initial velocity is 1.15 metres per second
Velocity after 7 seconds is 0.1 metres per second
Hope it helps!
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We have a body initially moving at a certain velocity, which we will call uu. Assuming constant acceleration of aa, this body is clearly slowing down, so a<0a<0 - if the body’s speed was constant, i.e. a=0a=0, the distance covered between time t=2 secondst=2 seconds to t=6 secondst=6 seconds would be 4 m4 m; and, if the speed was increasing, i.e. a>0a>0, the distance covered would have been more than 4 m4 m.
The obvious equation to use is:
s=ut+12at2s=ut+12at2, where ss is the distance travelled (in metres)
Now, let’s insert the information from the question into this equation.
A: 2=2u+1222a=2u+2a⇒u+a=12=2u+1222a=2u+2a⇒u+a=1
B: (2+2.2)=(2+4)u+12(2+4)2a(2+2.2)=(2+4)u+12(2+4)2a
⇒4.2=6u+18a⇒u+3a=0.7⇒4.2=6u+18a⇒u+3a=0.7
Subtracting A from B, we have: 2a=−0.3⇒a=−0.152a=−0.3⇒a=−0.15
Substituting this into A, we thus have: u−0.15=1⇒u=1.15u−0.15=1⇒u=1.15
So, what have we got so far?
The initial velocity of the object is 1.15 ms−11.15 ms−1
The object is accelerating at the rate of −0.15 m−2−0.15 m−2
Now, what do we want to find out?
From the question: What is [the object’s] velocity from the start to the end of the seventh second?
What does this mean? Well, there is more than one interpretation!
What seems (at least to me) the most likely meaning is that we have been asked to find the object’s initial velocity and the velocity at time t=7 secondst=7 seconds; in which case, the question should have said “What are the velocities at the start and after 7 seconds?”
Well, we have already calculated the initial velocity, so we’re half-way there!
The equation we need to determine the velocity, vv, at a certain time is:
v=u+atv=u+at
Feeding in the information we know: v=1.15+(−0.15)×7v=1.15+(−0.15)×7
=1.15−1.05=0.10=1.15−1.05=0.10
Thus the velocity after 7 seconds is 0.1 ms−10.1 ms−1
Answers:
Initial velocity is 1.15 metres per second
Velocity after 7 seconds is 0.1 metres per second
Hope it helps!
Give me more opportunity to help you by following me❤️☺️
KahkashaAnjum:
your answer is wrong.
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