Question

A car starts from rest and travels for 5.0 s with a uniform acceleration of +1.5m/s². The driver then applies the brakes, causing a uniform acceleration of -2.0 m/s². if the brakes are applied for 3.0 s, (a) how fast is the car going at the end of the braking period, and (b) how far has the car gone?

Answer 1

Answer:

Part a)

$v_2 = 1.5 m/s$

Part b)

$d = 39 m$

Explanation:

Part a)

Initial speed of the car = 0

acceleration of the car = 1.5 m/s^2

now at the end of t = 5 s the final speed of the car is given as

$v_f - v_i = at$

$v_1 - 0 = (1.5)(5)$

$v_1 = 7.5 m/s$

now after this car apply brakes which produce deceleration

so we have

$a = -2 m/s^2$

time to during which deceleration is applied is given by

$t = 3.0 s$

now we have

$v_f - v_i = at$

$v_2 - 7.5 = (-2)(3)$

$v_2 = 7.5 - 6 = 1.5 m/s$

Part b)

distance moved by car during the period of acceleration of the car is given as

$d = (\frac{v_f + v_i}{2})(time)$

$d_1 = (\frac{0 + 7.5}{2})(5)$

$d_1 = 18.75 m$

Now while car is decelerating then the distance moved by the car is given as

$d_2 = (\frac{7.5 + 1.5}{2})(3)$

$d_2 = 13.5 m$

now total distance moved by the car

$d = d_1 + d_2$

$d = 18.75 + 13.5$

$d = 32.25 m$

Answer 2

We use the equation of motions,

$s=ut+\frac{1}{2}at^2$                                             (A)

$v=u+at$                                                                  (B)

since a car starts from rest and travels for 5.0 s with a uniform acceleration of +1.5 m/s². therefore the distance traveled by the car for 5 s, from equation (A)

$s=0\times 5.0\s+\frac{1}{2}\times 1.5 m/s^2 \times (5.0\ s)^2 =18.75\ m$.

Now at the end of 5 s, the velocity of the car from (B),

$v=0+1.5\times 5 =7.5\ m/s$.

After the driver applied the brakes, the distance traveled by the car, again from equation (A)

$S=7.5\times 3.0s+\frac{1}{2} (-2.0 m/s^2)(3)^2=13.5\ m$.

At the end of the brakes applied, the velocity

$V=7.5+(-2\ m/s^2)\times3=15\ m/s.$.

The total distance traveled by the car,

$s+S=18.75\ m+13.5\ m=32.25\ m$