Question

The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration of −5.25 m/s2 for 4.15 s, making straight skid marks 62.5 m long, all the way to the tree. With what speed (in m/s) does the car then strike the tree? You may want to calculate the initial velocity of the car first. m/s (b) What If? If the car has the same initial velocity, and if the driver slams on the brakes at the same distance from the tree, then what would the acceleration need to be (in m/s2) so that the car narrowly avoids a collision? m/s2

Answer 1

Answer:

Part a)

Final speed of the car is

$v_f = 4.17 m/s$

Part b)

Acceleration of the car is

$a = -5.39 m/s^2$

Explanation:

As we know that car makes a skid of 62.5 m

here acceleration of the car is

$a = - 5.25 m/s^2$

now we have

$d = v_i t + \frac{1}{2}at^2$

$62.5 = v_i (4.15) + \frac{1}{2}(-5.25)(4.15^2)$

$v_i = 25.95 m/s$

Part a)

Speed of the car by which it will hit the tree

$v_f = v_i + at$

$v_f = 25.95 - (5.25)(4.15)$

$v_f = 4.17 m/s$

Part b)

Now if car will stop after travelling same distance which same initial speed

Then we can use kinematics

$v_f^2 - v_i^2 = 2 a d$

$0 - 25.95^2 = 2(a)(62.5)$

$a = -5.39 m/s^2$

Answer 2

Answer:

The car strikes the tree with a final speed of 4.165 m/s

The acceleration need to be of -5.19 m/seg2 to avoid collision by 0.5m

Explanation:

First we need to calculate the initial speed $V_{0}$

$x=V_{0} *t+\frac{1}{2} *a*(t^{2} )\\62.5m=V_{0} *4.15s+\frac{1}{2} *-5.25\frac{m}{s^{2} } *(4.15^{2} )\\V_{0}=25.953\frac{m}{s}$

Once we have the initial speed, we can isolate the final speed from following equation:

$V_{f} =V_{0} +a*t$  $V_{f}= 4.165 \frac{m}{s}$  

Then we can calculate the aceleration where the car stops 0.5 m before striking the tree.

To do that, we replace 62 m in the first formula, as follows:

$x=V_{0} *t+\frac{1}{2} *a*(t^{2} )\\62m=25.953\frac{m}{s}*4.15s+\frac{1}{2} *-a\frac{m}{s^{2} } *(4.15^{2} )\\a=-5.19\frac{m}{s^{2} }$