Question

A car starts from rest &Accelerates uniformly at the rate of 1m/s for 5 sec.It then maintains a constant velocity for 30 sec.Then brakes are applied & the car is uniformly retarded to rest in 10 sec.Find the maximum velocity attained by the car & the total distance travelled by it.

Answer 1

A car starts from rest & Accelerates uniformly at the rate of $1\mathrm{m/s}$ for 5sec. It then maintains a constant velocity for 30 sec. Then brakes are applied & the car is uniformly retarded to rest in 10 sec. Find the maximum velocity attained by the car &; the total distance travelled by it.

1. Acceleration:

$v = a t_{a} - \text{maximum velocity}$

$t_{a}$ – time of acceleration

$a$ – acceleration

Distance travelled:

$$l _ {a} = \frac {a t _ {a} ^ {2}}{2}$$

2. Uniform motion:

Distance travelled:

$$l _ {u} = v * t _ {u} = a t _ {a} t _ {u}$$

$t_{u}$ – time of uniform motion

3. Deceleration

$$v = a ^ {\prime} t _ {d}$$

$t_d$ – time of deceleration

$$a ^ {\prime} = \frac {v}{t _ {d}}$$

Distance travelled:

$$l _ {d} = \frac {a ^ {\prime} t _ {d} ^ {2}}{2} = \frac {v t _ {d}}{2} = \frac {a t _ {d} t _ {d}}{2}$$

Total distance:

$$\begin{array}{l} l = l _ {a} + l _ {u} + l _ {d} = \frac {a t _ {a} ^ {2}}{2} + a t _ {a} t _ {u} + \frac {a t _ {a} t _ {d}}{2} = a t _ {a} \left(\frac {t _ {a}}{2} + t _ {u} + \frac {t _ {d}}{2}\right) \\ = 1 \frac {m}{s ^ {2}} * 5 s \left(\frac {5}{2} s + 30 s + \frac {10}{2} s\right) = 5 \left(\frac {75}{2}\right) m = 187.5 m \\ \end{array}$$

Maximum velocity:

$$v = a t _ {a} = 1 \frac {m}{s ^ {2}} * 5 s = 5 \frac {m}{s}$$

Answer: $l = 187.5 \, \text{m}$, $v_{\text{max}} = 5 \frac{\text{m}}{\text{s}}$