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Sagot :
Answer:
s=ut+1/2at^2
where s=distance travelled
u=initial velocity
a=acceleration
t=time
ATQ, u=0m/s(as the body starts from rest, thus it's obvious that the initial velocity will be 0),t=20s, a=5m/s^2, s=?
Now put the value in the formula:
s=ut+1/2at^2
s=0(20s)+ 1/2(5m/s^2)(20s)^2
s=0+1/2(5m/s^2)(400s^2)
s=1/2(2000m)
s=1000m
Thus, distance travelled is 1000m
A body starts from rest and moves with a uniform acceleration of 5 m/s2, what is its velocity after 10 seconds?
A body moves from rest with a constant acceleration of 5m/s2. What is its instantaneous speed in m/s at the end of 10 seconds?
A body starts from rest if it is accelerated with 5m/s^2, then what is the distance covered by the body?
A body starts from first rest and reaches 5m/s travelling with uniform acceleration in a straight line for 2 seconds. What is the acceleration of the body?
What will be total distance travelled when a body starts from rest and with a uniform acceleration of 10 m/s^2 for 5 seconds? During the next 10 seconds, it moves with a uniform velocity.
Sum of all distances in each time interval (s) is given by
Distance traveled during 1st 1 second = 5 m
Distance traveled during 2nd 1 second = 10 m
Distance traveled during 3rd 1 second = 15 m
… … …
… … …
… … …
Distance traveled during 20th 1 second = 100 m
Now, by adding these distances, we get,
s = (5+10+15+…+100) m = 1050 m
For accuracy, we can use the formula from SUVRT equations given below:
s = (1/2)nrt, where number of ref. points (n) = 21, time (t) = 20 s, and praduryantar (r) = 5 m/s
Therefore,
s = (1/2)21*5*20 m = 1050 m.
This is the accurate result.
Initial velocity u=0,as body starts from rest.
Uniform acceleration a=5m/s^2
Time of travel t=20s
distance d=?
Equation of motion : d=ut+(1/2)at^2
d=0X20+(1/2)X5X400
d=1000m
A body starts from rest and moves with a uniform acceleration of 5 m/s2, what is its velocity after 10 seconds?
A bus starts from rest. If the acceleration is 2m/s^2, how would one find the distance travelled and the velocity after 2 seconds?
A body starts from rest and moves with uniform acceleration of 2m/s-2. What will be its velocity and displacement at the end of 5 seconds?
we know that S=ut+1/2 a*t^2
given that u=0, t=20s, a=5m/s^2
S=0*20+ (1/2)*5*(20)^2
S=1000m=1km
Body start from rest ,u=0,acc.=5m/s^2 time=20
S=ut+1/2at^2
S=0×20+1/2×5×20×20
S=1000m
The velocity at 20 second is 20*5 = 100 m /s .
Distance is average velocity * time = (100/2) * 20 = 1000 m
As
S=ut+1/2a*t^2
u=0
a=5 m/s^2,t=20s
S=0+1/2*5*20*20
S=1000m
Distance travelled by body is 1000m
The way to tackle a question like this is to sketch the v-t diagram.
In this case, it is a triangle starting at T=0 and u=0, ending at unknown final speed v at the 20s mark.
The slope of the line is 5m/s/s, and area under the line is the displacement you need.
You know how to find a slope and the area of a triangle …
A body starts from rest and moves with a uniform acceleration of 5 m/s2, what is its velocity after 10 seconds?
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