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

Byju's Answer

Standard IX

Physics

Derivation of Position-Velocity Relation by Graphical Method

The velocity ...

Question

The velocity V of a body moving along a straight line varies with time $t$ as $v = 2 t^{2} e^{- t}$ , where v is in m/s and $t$ in second . The acceleration of body is zero at $t$ =

0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Both (A) and (B)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D Both (A) and (B)
It is given that

$a = \frac{d v}{d t} = \frac{d}{d t} (2 t^{2} e^{- t})$

$= 2 [t^{2} e^{- 1} (- 1) + e^{- 1} 2 t]$

$= e^{- 1} (4 t - 2 t^{2})$
So, $a = 0$
$⟹ 4 t - 2 t^{2} = 0$
Or $2 t (2 - t) = 0$
$⟹ t = 0$ and $t = 2$

Suggest Corrections

Similar questions

Q. The velocity of a body moving along a straight varies with time as

V = t^{2} - 4 t m / s

, find the average acceleration of the body during an interval

t = 1 s e c

t = 3 s e c

Q. A particle is moving in straight line. The velocity v of the particle varies with time t as

v = t^{2} - 4 t

. then the distance traveled by the particle in

t = 0

t = 6 s

(where $t$ in second and v is in m/s )

Q. The speed of a particle moving in a circle of radius

r = 2 m

varies with time

t

v = t^{2}

where

t

is in second and

v

in m/s. Find the radial, tangential and net acceleration at

t = 2 s

Figure shows the acceleration – time ( $a - t$ ) graph of a body moving in a straight line. Which graph in figure shows the velocity – time ( $v - t$ ) graph of the motion of the body? (Assume that $x = 0$ and $v = 0$ at $t = 0$ )

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

)

Join BYJU'S Learning Program

The velocity V of a body moving along a straight line varies with time t as v=2t2e−t, where v is in m/s and t in second . The acceleration of body is zero at t=

The correct option is D Both (A) and (B)It is given that a=dvdt=ddt(2t2e−t) =2[t2e−1(−1)+e−12t] =e−1(4t−2t2) So, a=0 ⟹ 4t−2t2=0Or 2t(2−t)=0⟹ t=0 and t=2

The velocity V of a body moving along a straight line varies with time $t$ as $v = 2 t^{2} e^{- t}$ , where v is in m/s and $t$ in second . The acceleration of body is zero at $t$ =

The correct option is D Both (A) and (B)
It is given that

$a = \frac{d v}{d t} = \frac{d}{d t} (2 t^{2} e^{- t})$

$= 2 [t^{2} e^{- 1} (- 1) + e^{- 1} 2 t]$

$= e^{- 1} (4 t - 2 t^{2})$
So, $a = 0$
$⟹ 4 t - 2 t^{2} = 0$
Or $2 t (2 - t) = 0$
$⟹ t = 0$ and $t = 2$