Ncert Exercises & Solutions

3.3 A woman starts from her home at 9.00 am, walks with a speed of 5 km h^–1 on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by auto with a speed of 25 km h^–1. Choose suitable scales and plot the x-t graph of her motion.

From home to office: Time at which she leaves home for the office at 9 am.
speed = 5 km h^-1; distance = 2.5 km
Therefore, time taken to reach the office,
$t = \frac{d i s \tan c e}{s p e e d} = \frac{2.5}{5} = 0.5 h = 30 m i n .$
Thus, the time at which she reaches her office = 9.30 am.
Between 9.30 am to 5 pm, she remains in her office i.e. at a distance of 2.5 km from her
From office to home: Time at which she leaves her office = 5 pm.
Speed = 25 km h^-1, distance = 2.5 km
Therefore, time taken to reach the home,
$t = \frac{2.5}{25} = \frac{1}{10} h = 6 m i n$
Therefore, the time at which she reaches her home = 5.06 pm.

(x-t) graph of her motion will be as shown in the figure: $v_{s m o k e} = - 1500 k m / h r$

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