Question

A car accelerates from rest at a constant rate 'A' for sometime, after which it retards at a constant rate 'B' to come to rest. If the total time lapse is T seconds, evaluate the maximum velocity reached and total distance travelled , in terms of 'A', 'B' & 'T'?

Answer 1

Answer on Question #45896 – Physics - Mechanics | Kinematics | Dynamics

A car accelerates from rest at a constant rate 'A' for sometime, after which it retards at a constant rate 'B' to come to rest. If the total time lapse is T seconds, evaluate the maximum velocity reached and total distance travelled, in terms of 'A', 'B' & 'T'?

Solution:

Let $t$ and $t'$ are the time of acceleration and deceleration respectively. Let $v$ be maximum velocity.

$$t + t' = T$$ $$\frac{v}{t} = A, \quad \frac{v}{t'} = B$$

Thus, $t + t' = \frac{v}{A} + \frac{v}{B} = T$

$$T = v \left(\frac{A + B}{A B}\right)$$ $$v = \left(\frac{A B}{A + B}\right) T$$

Total distance traveled = area under $v - t$ graph

$$D = \frac{1}{2} (t + t') v = \frac{1}{2} T v = \frac{1}{2} \left(\frac{A B}{A + B}\right) T$$

**Answer**: maximum velocity: $v = \left(\frac{A B}{A + B}\right) T$

total distance: $D = \frac{1}{2} \left(\frac{A B}{A + B}\right) T$

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