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The position of a particle moving rectilinearly is given by x=t^3-3t^2-10. Find the distance travelled by the particle in the first 4 seconds starting from t=0.

The position of a particle moving rectilinearly is given by x=t^3-3t^2-10. Find the distance travelled by the particle in the first 4 seconds starting from t=0.

Grade:11

8 Answers

liopk
6 Points
6 years ago
position is given by x = t^3  – 3t^ 2 – 10 substitue 4 in place of t you will get 6 meters
 t^3 – 3t^2 – 10  = 4 ^3 – 3(4)^2 – 10 = (64 – 48) – 10 = 
6 meters 
therefore the distance travelled in first 4 seconds is 6 meters 
Shaunak Natarajan
35 Points
6 years ago
At T=0 he is at -10Also by derivative he`s direction of velocity changes at T=2,so putting t=2 we get x=-14 and then putting t=4 we get +4So total distance is (-10+14)+(4+14)=22
Soumya Ranjan Mohanty
117 Points
6 years ago
given: x=t^3-3t^2-10 --equation(1)
differentiating equation(1),we get v=3t^2-6t --equation(2)
 
at time t=0,from equation (1) the particle is at x=t^3-3t^2-10=0-0-10= -10. Now lets find out where the velocity is 0 i.e. its direction of motion is reversed(as it is given rectilinear motion). Setting equation(2) to 0, 3t^2-6t=0 =>3t(t-2)=0 or t=2 seconds. Therefore we have to calculate separately. At t=2, its position is x=t^3-3t^2-10= 2^3-3*2^2-10= 8-12-10= -14 . So from t=0 to t=2, distance travelled is x1=-10-(-14)=4 m. Now its position at t=4 ,x=x=t^3-3t^2-10=4^3-3*4^2-10=64-48-10=6, so from t=2 to t=4, it travelled x2=6-(-14)=6+14=20m.Therefore total distance travelled by the particle in 4 seconds starting from t=0 is x1+x2=4+20= 24m.
So ans is 24m.
 
Thanks,
Soumya Ranjan Mohanty
vivek jaiswal
37 Points
6 years ago
Since x=t^3-3t^2-10=} dx/dt=v=3t^2-6tTurned where v=03t(t-2)=0t=2 where turneddx=vdtX=intrigation(0 to 2){3t^2-6t} - intrigation (2 to 4){3t^2-6t}X=[3t^3/3 - 6t^2/2](0 to 2) -[3t^3/3 - 6t^2/2](2 to 4)X=-24mSince distance is positive answer is 24m
Amarjit
105 Points
6 years ago
On differentiation we see that that the particle has changed its direction at t=2 . At t= 0 we find that its position is at x=-10 and at t=2 we see that it is at x=-14 and at t=4 we see that it is at x=6. Hence total distance travelled= 4+14+6=24 units.
Rudra
14 Points
6 years ago
At t=0, x= -10 unitswhere at t=4, x=6 units.Since the distance is from -10 units to 6 units.Total distannce=|-10|units+|6 units|= 16 units.
Gitanjali Rout
184 Points
5 years ago
at time t=0,from equation (1) the particle is at x=t^3-3t^2-10=0-0-10= -10. Now lets find out where the velocity is 0 i.e. its direction of motion is reversed(as it is given rectilinear motion). Setting equation(2) to 0, 3t^2-6t=0 =>3t(t-2)=0 or t=2 seconds. Therefore we have to calculate separately. At t=2, its position is x=t^3-3t^2-10= 2^3-3*2^2-10= 8-12-10= -14 . So from t=0 to t=2, distance travelled is x1=-10-(-14)=4 m. Now its position at t=4 ,x=x=t^3-3t^2-10=4^3-3*4^2-10=64-48-10=6, so from t=2 to t=4, it travelled x2=6-(-14)=6+14=20m.Therefore total distance travelled by the particle in 4 seconds starting from t=0 is x1+x2=4+20= 24m.
So ans is 24m.
o 0, 3t^2-6t=0 =>3t(t-2)=0 or t=2 seconds. Therefore we have to calculate separately. At t=2, its position is x=t^3-3t^2-10= 2^3-3*2^2-10= 8-12-10= -14 . So from t=0 to t=2, distance travelled is x1=-10-(-14)=4 m. Now its position at t=4 ,x=x=t^3-3t^2-10=4^3-3*4^2-10=64-48-10=6, so from t=2 to t=4, it travelled x2=6-(-14)=6+14=20m.Therefore total distance travelled by the particle in 4 seconds starting from t=0 is x1+x2=4+20= 24m.So ans is 24m
Ashwin Sheoran
44 Points
5 years ago
We cab calculate the distance by substituting t by 4 we can find velocity by differentiation and  accelarationby double differentiation. Substituting the value of t by 4 we will find that the distance travelled in the first four seconds will be 6
 
 
 

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