Acceleration of a body is given by dv dt =4-2v. The body started … - Vidyadip

MCQ

Acceleration of a body is given by $\frac{\text{dv}}{\text{dt}}=4-2\text{v}.$ The body started from rest $t = 0.$ Then:

✓
The acceleration of the body is initially $+ ve$ but after some time it becomes negative.
B
The body will ultimately acquire a constant velocity after a long time.
C
The body will finally come rest.
D
The acceleration will finally attain a constant non-zero value.

✓

Answer

Correct option: A.

The acceleration of the body is initially $+ ve$ but after some time it becomes negative.

Given that,
Acceleration, $\text{a}=\frac{\text{dv}}{\text{dt}}=4-2\text{t}.$
At $t = 0s, a = 4\ m/ s^2$
At $t = 1s, a =2\ m/ s^2$
At $t = 2s, a = 0$
At $t = 3s, a = 4 − 6 = −2\ m/ s^2$
Hence, the acceleration of the body is initially positive but after some time it becomes negative.

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