Question

The velocity (v) of a particle starting from rest increases linearly with time (t) as $ v = 4t $, where v is in $ ms^{-1} $ and t is in second. The distance covered by the particle in the first 4 seconds is

• A. $16 \, m$
• B. $32 \, m$
• C. $8 \, m$
• D. $64 \, m$

Hint:

For motion with variable velocity, the distance covered is the definite integral of the velocity function over the time interval.

Answer 1

The Correct Option is B

Step 1: Identify the velocity function.
The velocity is given by $ v(t) = 4t $.
Step 2: Recall the relationship between velocity and distance.
Distance is the integral of velocity with respect to time: $ s = \int v(t) \, dt $.
Step 3: Set up the definite integral for the given time interval.
We need to find the distance covered from $ t = 0 $ to $ t = 4 $ seconds: $$ s = \int_{0}^{4} 4t \, dt $$
Step 4: Evaluate the definite integral.
$$ s = 4 \int_{0}^{4} t \, dt = 4 \left[ \frac{t^2}{2} \right]_{0}^{4} $$ $$ s = 4 \left( \frac{(4)^2}{2} - \frac{(0)^2}{2} \right) = 4 \left( \frac{16}{2} - 0 \right) = 4 (8) = 32 $$
Step 5: State the result with units.
The distance covered is $ 32 \, m $.
Thus, the distance covered by the particle in the first 4 seconds is $ \boxed{32 \, m} $.