A car starts from rest and moves with uniform acceleration '$a$' on a straight road from time $t = 0$ to $t =

Question

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:

Accepted Answer

1.  **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.
2.  **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 $v_{max} = 0 + aT = aT$ .
3.  **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 = $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes t_{total} 	imes v_{max}$ .
4.  **Calculate Average Speed:** Average speed is defined as the total distance divided by the total time.
    *   $v_{avg} = \frac{S}{t_{total}} = \frac{\frac{1}{2} 	imes t_{total} 	imes v_{max}}{t_{total}} = \frac{v_{max}}{2}$.
5.  **Substitute Value:** Substituting $v_{max} = aT$, we get:
    *   $v_{avg} = \frac{aT}{2}$.

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