Thread: particles

A particle is moving along a straight trajectory. The distance covered by a particle is described by the function S(t) = 2t^3 + 5t^2 , where t > 0 is the time.

a)What is the distance covered by the particle by time t = 4 ?

b)Find approximate value of time t when the speed of the particle is 84

c)Find a function that describes the acceleration of this particle.

d)Find the acceleration of the particle at t = 5

e)Find the time when the acceleration of the particle is 46

a)What is the distance covered by the particle by time t = 4 ?

the velocity is the derivative of the displacement. Find the times at which the particle is stationary between t=0 and t=4 (or establish that there are none), then add up the displacements between consecutive stationary points and t=0 and t=4.

b)Find approximate value of time t when the speed of the particle is 84

Speed is the magnitude of velocity. velocity is the derivative of displacement with respect to t.

c)Find a function that describes the acceleration of this particle.

acceleration is the derivative of velocity with respect to t.

You will probably be able to do d and e when you have done c.

A particle is moving along a straight trajectory. The distance covered by a particle is described by the function S(t) = 2t^3 + 5t^2 , where t > 0 is the time.

a)What is the distance covered by the particle by time t = 4 ?
diiferentiate and area under graph is distance
therefore dS/dt=6t^2 + 10t
sub t=4
we get total distance as 136
Ans:136

b)Find approximate value of time t when the speed of the particle is 84
using the form 6t^2 + 10t - 84 = 0
reduce it to 3t^2 + 5t - 42
than a=3 , b= 5 , c= -42

put it in formula

-b+/- Squart(b^2 -4ac)/2a

we get 1.68878secconds
Ans:1.68878

c)Find a function that describes the acceleration of this particle.

differentiate velocity

ans: 12t +10

d)Find the acceleration of the particle at t = 5

12(5) + 10 = 70

Ans:70



e)Find the time when the acceleration of the particle is 46

46 = 12t + 10
therefore t= 3 sec

plse assist to check my answer...thanks.

The answer for a) is

The rest is correct.

if the 1st ans is correct ..than my second answer should be wrong rite
because

as u mentioned from the first ans::::::

84=2t^3 + 5t^2 for part b) and than use long devision to solve rite ??

s(t) is the distance covered.
To know the distance covered at t=4 you just substitute t by 4 in s(t).
To find the time at which the speed is 84 you do as you did

You solve with respect to t.

It seems I made a mistake too. You show be ok now. n+1 heads are better n heads.

a)What is the distance covered by the particle by time t = 4 ?
diiferentiate and area under graph is distance
therefore dS/dt=6t^2 + 10t

Then set and find the values of t that satisfy this. The reason we do this is so that we know when the particle changes direction. For example, if Billy goes 100m right in the first 2 minutes and then 50m left in the next 3 minutes, the distance covered for the 5 minutes is 150m, but evaluating S(5) would give only his current position, 50m.

Once we know when the particle changes direction, we can get the distance travelled by adding up the distances it travelled between changing direction. Since we know it didn't change direction in between changes in direction, we can find these distances by subtracting the positions.

b)Find approximate value of time t when the speed of the particle is 84
using the form 6t^2 + 10t - 84 = 0
reduce it to 3t^2 + 5t - 42
than a=3 , b= 5 , c= -42

put it in formula

-b+/- Squart(b^2 -4ac)/2a

we get 1.68878secconds
Ans:1.68878

Your working and formula are correct (although the formula is missing some brackets: ( -b+/- Squart(b^2 -4ac))/2a) but the answer is not. Try entering it into your calculator again.

c)Find a function that describes the acceleration of this particle.

differentiate velocity

ans: 12t +10

d)Find the acceleration of the particle at t = 5

12(5) + 10 = 70

Ans:70



e)Find the time when the acceleration of the particle is 46

46 = 12t + 10
therefore t= 3 sec

all right. Well done.