JEE MAIN - Physics (2003 - No. 54)
A car, moving with a speed of 50 km/hr, can be stopped by brakes after at least 6 m. If the
same car is moving at a speed of 100 km/hr, the minimum stopping distance is
12 m
18 m
24 m
6 m
Explanation
For case 1 :
u = 50 km/hr = $${{50 \times 1000} \over {3600}}$$ m/s = $${{125} \over 9}$$ m/s, v = 0, s = 6 m, $$a$$ = ?
$$\therefore$$ 02 = u2 + 2$$a$$s
$$ \Rightarrow $$ $$a = - {{{u^2}} \over {2s}}$$
$$ \Rightarrow $$ $$a = - {{{{\left( {{{125} \over 9}} \right)}^2}} \over {2 \times 6}}$$ = $$-$$16 m/s2
For case 2 :
u = 100 km/hr = $${{100 \times 1000} \over {3600}}$$ m/s = $${{250} \over 9}$$ m/s, v = 0, $$a$$ = $$-$$16, s = ?
$$\therefore$$ 02 = u2 + 2$$a$$s
$$ \Rightarrow $$ $$s = - {{{u^2}} \over {2a}}$$
$$ \Rightarrow $$ $$s = - {{{{\left( {{{250} \over 9}} \right)}^2}} \over {2 \times -16}}$$ = 24 m
u = 50 km/hr = $${{50 \times 1000} \over {3600}}$$ m/s = $${{125} \over 9}$$ m/s, v = 0, s = 6 m, $$a$$ = ?
$$\therefore$$ 02 = u2 + 2$$a$$s
$$ \Rightarrow $$ $$a = - {{{u^2}} \over {2s}}$$
$$ \Rightarrow $$ $$a = - {{{{\left( {{{125} \over 9}} \right)}^2}} \over {2 \times 6}}$$ = $$-$$16 m/s2
For case 2 :
u = 100 km/hr = $${{100 \times 1000} \over {3600}}$$ m/s = $${{250} \over 9}$$ m/s, v = 0, $$a$$ = $$-$$16, s = ?
$$\therefore$$ 02 = u2 + 2$$a$$s
$$ \Rightarrow $$ $$s = - {{{u^2}} \over {2a}}$$
$$ \Rightarrow $$ $$s = - {{{{\left( {{{250} \over 9}} \right)}^2}} \over {2 \times -16}}$$ = 24 m
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