Following table gives the displacement of a car at different instants of time.
Times (s) 0 1 2 3 4
Displacement (m) 0 5 10 15 20
What will be the displacement of car at (i) 25 s and (ii) 45 s ?

(i) 25 m
(ii) 45 m

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(i) 625 m
(ii) 125 m

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(i) 125 m
(ii) 225 m

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(i) 45 m
(ii) 25 m

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Solution

The correct option is C (i) 125 m
(ii) 225 m
From the table we can see that the car is moving with a constant velocity of $5 m s^{- 1}$
i.e.
Distance traveled after:
$25 s; d = 25 \times 5 = 125 m$
$45 s; d = 45 \times 5 = 225 m$
Hence option C is correct.

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Similar questions

Following table gives the displacement of a car at different instants of time.

$\begin{matrix} T i m e (s) & 0 & 1 & 2 & 3 & 4 D i s p l a c e m e n t (m) & 0 & 5 & 10 & 15 & 20 \end{matrix}$

(a) Draw the displacement-time sketch and find the average velocity of car.

(b) What will be the displacement of car at

(i) 2.5 s and (ii) 4.5 s ?

\frac{- 12}{5} \times \frac{3}{4} = \frac{3}{4} \times \frac{- 12}{5}

\frac{- 8}{15} \times 1 = \frac{- 8}{15}

(\frac{- 2}{3} \times \frac{7}{8}) \times \frac{- 5}{7} = \frac{- 2}{3} \times (\frac{7}{8} \times \frac{- 5}{7})

\frac{- 2}{3} \times 0 = 0

\frac{2}{5} \times (\frac{- 4}{5} + \frac{- 3}{10}) = (\frac{2}{5} \times \frac{- 4}{5}) + (\frac{2}{5} \times \frac{- 3}{10})

Times (s)	0	1	2	3	4
Displacement (m)	0	5	10	15	20

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