For a given velocity-time (v-t) graph, the total time is 10 s. Given that its maximum velocity during motion is 3 m / s and α=45°, then at what time (in s) the particle is moving with uniform velocity?

Question

For a given velocity-time (v-t) graph, the total time is 10 s. Given that its maximum velocity during motion is 3 m / s and α=45°, then at what time (in s) the particle is moving with uniform velocity? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

t tan 45°=3 m / s ⇒ t=3 s So, 10-2 t=4 s

Q. For a given velocity-time $(v - t)$ graph, the total time is $10 s$ . Given that its maximum velocity during motion is $3 m / s$ and $α = 4 5^{\circ}$ , then at what time (in $s$ ) the particle is moving with uniform velocity?

$t tan 4 5^{\circ} = 3 m / s$
$\Rightarrow t = 3 s$
So, $10 - 2 t = 4 s$

Q. For a given velocity-time (v−t) graph, the total time is 10s. Given that its maximum velocity during motion is 3m/s and α=45∘, then at what time (in s) the particle is moving with uniform velocity?

ttan45∘=3m/s ⇒t=3s So, 10−2t=4s

Q. For a given velocity-time $(v - t)$ graph, the total time is $10 s$ . Given that its maximum velocity during motion is $3 m / s$ and $α = 4 5^{\circ}$ , then at what time (in $s$ ) the particle is moving with uniform velocity?

$t tan 4 5^{\circ} = 3 m / s$
$\Rightarrow t = 3 s$
So, $10 - 2 t = 4 s$