The given graph is (1/ text acceleration (a)) vs velocity (v) graph. If the time interval during which velocity changes from 2 m / s to 4 m / s is given by Δ t seconds, then find the value of 2 Δ t.

Question

The given graph is (1/ text acceleration (a)) vs velocity (v) graph. If the time interval during which velocity changes from 2 m / s to 4 m / s is given by Δ t seconds, then find the value of 2 Δ t. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

(1/a)=(1/4) v (d t/d v)=(1/4) v ; ∫ limitst1t2 d t=∫ limits24 (v/4) d v t2-t1=(1/4) × (1/2)[16-4]=(3/2) s

Q. The given graph is $\frac{1}{acceleration ( a )}$ vs velocity $(v)$ graph. If the time interval during which velocity changes from $2 m / s$ to $4 m / s$ is given by $Δ t$ seconds, then find the value of $2Δ t$ .

$\frac{1}{a} = \frac{1}{4} v$
$\frac{d t}{d v} = \frac{1}{4} v; t_{1} \int t_{2} d t = 2 \int 4 \frac{v}{4} d v$
$t_{2} - t_{1} = \frac{1}{4} \times \frac{1}{2} [16 - 4] = \frac{3}{2} s$

Q. The given graph is acceleration (a)1​ vs velocity (v) graph. If the time interval during which velocity changes from 2m/s to 4m/s is given by Δt seconds, then find the value of 2Δt.

a1​=41​v dvdt​=41​v;t1​∫t2​​dt=2∫4​4v​dv t2​−t1​=41​×21​[16−4]=23​s

Q. The given graph is $\frac{1}{acceleration ( a )}$ vs velocity $(v)$ graph. If the time interval during which velocity changes from $2 m / s$ to $4 m / s$ is given by $Δ t$ seconds, then find the value of $2Δ t$ .

$\frac{1}{a} = \frac{1}{4} v$
$\frac{d t}{d v} = \frac{1}{4} v; t_{1} \int t_{2} d t = 2 \int 4 \frac{v}{4} d v$
$t_{2} - t_{1} = \frac{1}{4} \times \frac{1}{2} [16 - 4] = \frac{3}{2} s$