A car starts from rest and travels for 5.0s with a uniform acceleration of +1.8

Question

A car starts from rest and travels for 5.0s with a uniform acceleration of +1.8 m/s2. Thedriver then applies the brakes, causing a uniform acceleration of-3.0m/s2. If the brakes are applied for 1.0 s, determine each of the following. (a) How fast is the car going at the end of thebraking period?
1 m/s
(b) How far has it gone?
2 m
Part b A parachutist with a camera descends in free fall at a speedof 11 m/s. The parachutist releases thecamera at an altitude of 140 m.

(a) How long does it take the camera to reach theground?
1 s
(b) What is the velocity of the camera just before it hits theground?
2 m/s

(a) How fast is the car going at the end of thebraking period?
1 m/s
(b) How far has it gone?
2 m
Part b A parachutist with a camera descends in free fall at a speedof 11 m/s. The parachutist releases thecamera at an altitude of 140 m.

(a) How long does it take the camera to reach theground?
1 s
(b) What is the velocity of the camera just before it hits theground?
2 m/s

Explanation / Answer

part a----------- u=0 a=1.8 t=5 v=u+a*t v=0+1.8*5 v=9 m/s now a=-3 u=9 v=velocity after breaking period t=1 v=u-a*t v=9-3*1 v= 6 m/s (ans ) during first 5 s the car traveled s1 m s1=u*t+(1/2)*a*t2s1=0+(1/2)*1.8*52 s1=22.5 m during the next 1 s the car traveled s2 m s2=9*1-(1/2)*3*12 s2=9-1.5 s2=7.5 m total distance traveled from rest=22.5+7.5=30 m part b---------- initial velocity of the camera=11 m/s towards the ground h=140 m velocity when it hit the ground=v v2=u2+2*g*h v2=112+2*9.8*140 v=53.53 m/s (ans) time taken to reach the ground=t v=u+g*t 53.53=11+9.8*t t=4.34 s (ans)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.