Answer:
a. 17.34 s. b. 104.51 m c. -7.16 m/s²
Explanation:
a. The distance, d moved by the car at constant speed v = 6.00 m/s in time t which they collide is d = vt. The distance d' moved by the bus from its initial position 70 m away from the car with a constant velocity u = 28.0 m/s and deceleration a = -3.00 m/s² is d' = 70 + ut + 1/2at².
At collision, s = d
vt = 70 + ut + 1/2at²
substituting the values of the variables, we have
6t = 70 + 28t + 1/2(-3)t²
6t = 70 + 28t - 1.5t²
collecting like terms, we have
1.5t² + 6t - 28t - 70 = 0
1.5t² - 22t - 70 = 0
using the quadratic formula to find t, we have
We take the positive value.
So t = 17.34 s.
So, 17.34 s elapses before the bus and car collide.
b. Since the distance covered by the bus is d' = 70 + ut + 1/2at², this is how far down the road the collision occurs.
Substituting the values of the variables, we have
d' = 70 + 28 m/s × 17.34 s + 1/2(-3 m/s²)(17.34)²
d' = 70 + 485.52 - 451.01
d'= 104.51 m
c. To just avoid the collision, the bus must decelerate in the distance of 104.51 m from the car to the speed of the car. So using
v² = u² + 2ad' where v = 6.00 m/s, u = 28.0 m/s and d' = 104.51 m
the acceleration is thus
a = (v² - u²)/d'
a = ((6 m/s)² - (28 m/s)²)/104.51 m
a = (36 m²/s² - 784 m²/s²)/104.51 m
a = -748 m²/s² ÷ 104.51 m
a = -7.16 m/s²
