Question

On a straight stretch track a toy car is set rolling with an initial position of 16.0 cm, an initial velocity of-3.1 cm/s, and a constant acceleration of 2.10 cm/s​2​. At the same moment, another toy car is set rollingon an adjacent track with an initial position of 11.5 cm, an initial velocity of 5.40 cm/s, and a constantzero acceleration.
a.At what time, if any, do the two cars have equal speeds?
b.What are their speeds at that time?
c.At what time(s), if any, do the cars pass each other?
d.What are their locations at that time?

Answer 1

Answer:

Part a)

the time after which both cars will have same speed is t = 4.05 s

Part b)

$v_a = v_b = 5.4 cm/s$

Part c)

t = 5.12 s

Part d)

$x_a = x_b = 39.15 cm$

Explanation:

Part a)

As we know that Car A is moving with initial speed 3.1 cm/s with deceleration 2.10 cm/s/s while car B is moving with constant 5.40 cm/s speed

Now if the speed of Car A and Car B will be same after some time

So we will have

$v_a = v_b$

$-3.10 + (2.10) t = 5.40$

$t = 4.05 s$

so the time after which both cars will have same speed is t = 4.05 s

Part b)

After this time when both cars have same speed

this speed will be same as that of speed of Car B as it is moving at constant speed

so we have

$v_a = v_b = 5.4 cm/s$

Part c)

if final position of both cars will be same

then we will have

$x_a = x_i + v_a t + \frac{1}{2}at^2$

$x_a = 16 + (-3.1) t + \frac{1}{2}(2.10)t^2$

also we have

$x_b = x_i + vt$

$x_b = 11.5 + 5.40 t$

now we have

$16 - 3.1 t + 1.05 t^2 = 11.5 + 5.40 t$

$1.05 t^2 - 8.50t + 16 = 0$

t = 5.12 s

Part d)

Final positions of both the particles is given as

$x_b = 11.5 + 5.40 t$

$x_b = 11.5 + (5.40)(5.12)$

$x_b = 39.15 cm$