Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A car starts from rest and moves with uniform acceleration ' $a$ ' on a straight road from time $t = 0$ to $t = T$ . After that, a constant deceleration brings it to rest. In this process the average speed of the car is:

$\frac{aT}{4}$

$\frac{3aT}{2}$

$\frac{aT}{2}$

$aT$

Step-by-Step Solution

Analyze the Motion: The motion consists of two phases: accelerating from rest to a maximum velocity ( $v_{max}$ ) and then decelerating from $v_{max}$ to rest.
Maximum Velocity: During the acceleration phase (from $t=0$ to $t=T$ ), the initial velocity $u=0$ and acceleration is $a$ . Using the equation $v = u + at$ , the maximum velocity reached is $v_{max} = 0 + aT = aT$ .
Velocity-Time Graph Method: The velocity-time graph for this motion is a triangle starting from the origin, peaking at $v_{max}$ at time $T$ , and dropping to zero at total time $t_{total}$ .

The total distance covered ( $S$ ) is the area under the v-t graph: Area = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times t_{total} \times v_{max}$ .

Calculate Average Speed: Average speed is defined as the total distance divided by the total time.

$v_{avg} = \frac{S}{t_{total}} = \frac{\frac{1}{2} \times t_{total} \times v_{max}}{t_{total}} = \frac{v_{max}}{2}$ .

Substitute Value: Substituting $v_{max} = aT$ , we get:

$v_{avg} = \frac{aT}{2}$ .

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