A car is traveling along a straight road at a velocity of +31.0 m/s when its eng

Question

A car is traveling along a straight road at a velocity of +31.0 m/s when its engine cuts out. For the next ten seconds, the car slows down, and its average acceleration is a1. For the next five seconds, the car slows down further, and its average acceleration is a2. The velocity of the car at the end of the fifteen-second period is +21.0 m/s. The ratio of the average acceleration values is a1/a2 = 1.60. Find the velocity of the car at the end of the initial ten-second interval. m/s Please show all work

Explanation / Answer

As given motion is straight line

newtons equation u = v+at can be used

let v1 be velocity after 10 seconds an v2 is after 15 seconds

u=31m/s

for first 10sec a=a1 and after a2

31 = v1 +a1*10 (1)

now u becomes v1

v1=v2+a2 *5

v1 = 21+a2*5 (2)

a1=1.6a2 (3)

using eqn 1,2 and3

31 =21 +1.6*a2*10 +5a2

a2 = 10/21

so v1 = 21+10/21 *5

=491/21

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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