The motion of a car along Y axis is given by the relation y= t^3-6t^2+9t+5. Y is in metres and t is in sec.calculate the position,velocity and acceleration at t=5 sec and distance travelled during 0 to 5 sec

Open in App

Solution

Displacement y is given

v = velocity

a = acceleration

y= t³- 6t² + 9t +5

dy/dt = d/dt ( t³ ) - d/dt (6t²) + d/dt (9t) + d/dt (5)

We know dy/dt = v

So , v = 3t² - 12t + 9 + 0

dv/dt = d/dt ( 3t²) - d/dt ( 12t ) + d/dt (9)

we also know dv/dt = a

a = 6t - 12

Putting t= 5 s.

acceleration = (6*5 - 12) m/s²

= (30 - 12) m/s²

= 18 m/s²

And displacement is ( put t = 5s in y= t³ - 6t²+9t + 5

displacement= [(5)³ - 6.(5)² + 9 . (5) + 5] m

=[125 - 150 + 45 + 5] m

= 25 m

Suggest Corrections

Similar questions

Q. A particle is moving along the path given by

y = \frac{c}{6 t}

( where c is a positive constant). The relation between the acceleration (a) and the velocity (v) of the particle at

t = 5

sec. is

A particle is moving along the path given by y = $\frac{c}{6}$ $t^{6}$ (where c is positive constant). The relation between the acceleration (a) and the velocity (v) of the particle at t=5 sec. Is

Q. The position of a particle travelling along x axis is given by

x_{1} = t^{3} - 9 t^{2} + 6 t

where

x_{t}

is in

c m

and

t

is in second. Then

Q. The position of particle travelling along x-axis is given by

x_{t} = t^{3} - 9 t^{2} + 6 t

where

x_{t}

is in

cm

and

t

is in second. Then–

The displacement (in meter) of a particle moving along the x-axis is given $x = 18 t + 5 t^{2} .$ Calculate

(A) The instantaneous velocity at t = 2 sec

(B) Average velocity between t = 2 sec and t = 3 sec

Join BYJU'S Learning Program