Question

A car traveling at constant velocity of 20m/s passes a stationary police car and 15 seconds later the police car accelerates uniformly at 4m/s to reach a speed of 30m/s with which it continues. a) draw a velocity time graph. b) find where the second car overtakes the first car. c) find the distance up to which the second car overtakes the first car

Answer 1

Solution.

$v_c=20m/s;$

$t_c=15s;$

$v_{p0}=0m/s;$

$v_p=30m/s;$

$a_p=4m/s;$

$a)$

$b) S_c=v_c\sdot t_c=20\sdot15=300m;$

$a_p=\dfrac{v_p-v_{p0}}{t_p}\implies t_p=\dfrac{v_p-v_{p0}}{a_p};$

$t_p=\dfrac{30-0}{4}=7.5s;$

$x_{c0}=S_c+v_ct_p=300+20\sdot7.5=450m;$

$x_c=450+20t;$

$x_p=30t;$

$x_c=x_p;$

$450+20t=30t;$

$10t=450;$

$t=45s;$

$c)x_c=450+20t=450+20\sdot45=1350m;$

Answer: $b)t=45s;$

$c) x_c=1350m.$