You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Finding Average Velocity

A particle mo...

Question

A particle moves along a horizontal path, such that its velocity is given by $v = (3 t^{2} - 6 t)$ m/s, where t is the time in seconds. If it is initially located at the origin O, determine the distance travelled by the particale in time interval from t = 0 to t= 3.5 s and the particle's average velocity and average speed druing the same time interval.

Open in App

Solution

$v e l o c i t y o f t h e p a r t i c l e b e c o m e s z e r o a t t = 2 s a n d t h e d i r e c t i o n i s c h a n g e d a f t e r t h a t . d i s t a n c e = ∣ ∣ \int_{0}^{2} (3 t^{2} - 6 t) d t ∣ ∣ + ∣ ∣ \int_{2}^{3.5} (3 t^{2} - 6 t) d t ∣ ∣ = | - 4 | + | 6.125 | = 10.125 m$

Suggest Corrections

Similar questions

Q. A particle moves along a straight line and its velocity depends on time as

v = 6 t - 3 t^{2}

where

v

is in

m/s

and

t

in seconds. Find the average speed for first 4 seconds. (in

m/s

)

Q. A body moves in a straight line along

Y -

axis. Its distance

y

(in metre) from the origin is given by

y = 8 t - 3 t^{2}

(

t

in seconds). The average speed in the time interval from

t = 0

second to

t = 2

second is

A particle starts moving rectilinearly at time t = 0 such that its velocity ‘v’ changes with time ‘t’ according to the equation $v = t^{2} - t$ where t is in seconds and v in m/s. Find the time interval for which the particle retards.

Q. A body moves in a straight line along Y-axis. Its distance

y

(in metre) from the origin is given by

y = 8 t - 3 t^{2}

. The average speed in the time interval from

t = 0

second to

t = 1

second is

Q. A particle moves along x-axis and its acceleration at any time

t

a = 2 s i n (π t)

, where

t

is in seconds and

a

is in

m / s^{2}

. The initial velocity of particle (at time

t = 0

) is

u = 0

. The distance travelled (in meters) by the particle from time

t = 0

t = 1 s

will be

Join BYJU'S Learning Program

velocityoftheparticlebecomeszeroatt=2sandthedirectionischangedafterthat.distance=∣∣∫20(3t2−6t)dt∣∣+∣∣∫3.52(3t2−6t)dt∣∣=|−4|+|6.125|=10.125m

$v e l o c i t y o f t h e p a r t i c l e b e c o m e s z e r o a t t = 2 s a n d t h e d i r e c t i o n i s c h a n g e d a f t e r t h a t . d i s t a n c e = ∣ ∣ \int_{0}^{2} (3 t^{2} - 6 t) d t ∣ ∣ + ∣ ∣ \int_{2}^{3.5} (3 t^{2} - 6 t) d t ∣ ∣ = | - 4 | + | 6.125 | = 10.125 m$