Question

A car starts from rest and accelerates uniformly to a speed of $50 m s^{- 1}$ over a distance of $250 m$ .

(a) Find the acceleration and time taken by the car to cover this distance.

(b) Then, the speed of car is increased to $60 m s^{- 1}$ within $10 s$ if the acceleration is further increased. Find this acceleration and the distance travelled by car in $10 s$ .

(c) To stop the car in $6 s$ , the brakes are applied. Find the distance travelled by the car during retardation.

Accepted Answer

Solution for (a):

Given,

A car accelerates uniformly with

Initial speed, $u = 0$ ,

Final speed, $v_{1} = 50 m s^{- 1}$ and

Displacement, $10 s$ .

Using the third equation of motion under uniform acceleration,

$v_{1}^{2} = u^{2} + 2 a_{1} s_{1}$

$\Rightarrow 50^{2} = 0 + 2 \times a_{1} \times 250$

$\Rightarrow a_{1} = 5 m s^{- 2}$ .

Now, by first equation of motion,

$v_{1} = u + a_{1} t_{1}$

$\Rightarrow 50 = 0 + 5 t_{1}$

$\Rightarrow t_{1} = 10 s$ .

Therefore, the acceleration during this motion is $5 m s^{- 2}$ and the time taken during motion is $10 s$ .

Solution for (b):

For motion of car, after covering the distance of $250 m$ .

Initial speed, $v_{1} =$ $50 m s^{- 1}$ ,

Final speed, $v_{2} = 60 m s^{- 1}$ and

time $t_{2} = 10 s$ .

Using first equation of motion, $v_{2} = v_{1} + a_{2} t_{2}$ ,

$\Rightarrow 60 = 50 + a_{2} \times 10$ ,

$\Rightarrow a_{2} = 1 m s^{- 2}$ .

Also, by second equation of motion,

$s_{2} = v_{1} t + \frac{1}{2} a t_{2}^{2} = (50 \times 10) + \frac{1}{2} \times 1 \times 10^{2} = 550 m$ .

Solution for (c):

Here, we take initial velocity as $60 m s^{- 1}$ .

So, $v_{2}$ = $60 m s^{- 1}$ , $v$ = $0$ and time, $t_{3}$ = $6 s$ .

Using first equation of motion, $v = v_{2} + a_{3} t_{3}$ ,

$\Rightarrow 0 = 60 + a_{3} \times 6$ ,

$\Rightarrow a_{3} = - 10 m s^{- 2}$ .

Now, using second equation of motion, $s_{3} = v_{2} t + \frac{1}{2} a_{3} t^{2}$ ,

$s_{3} = (60 \times 6) + \frac{1}{2} \times - 10 \times 6^{2} = 180 m$ .

So, the distance travelled during retardation is $180 m$ .

B M Sharma Solutions for Chapter: Kinematics I, Exercise 24: CONCEPT APPLICATION EXERCISE 4.2

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Kinematics I, Exercise 24: CONCEPT APPLICATION EXERCISE 4.2

Questions from B M Sharma Solutions for Chapter: Kinematics I, Exercise 24: CONCEPT APPLICATION EXERCISE 4.2 with Hints & Solutions

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