An object undergoes constant acceleration after starting from rest and then travels 5 m in the first second…

Question

An object undergoes constant acceleration after starting from rest and then travels 5 m in the first second. Determine how far it will go in the next second.

ScanSolve AI Tutor · Accepted Answer

Exercise 1 Step 1: Formula for average speed v_avg = total distancetotal time. Total distance s = 50 km. Total time t = 30 min = 30/60 h = 0.5 h. Substitute: v_avg = 50/0.5 km/h. Simplify: v_avg = 100 km/h. 100 km/h Exercise 2 Step 1: The aircraft flies in a straight line due north, so magnitude of displacement equals distance traveled. Magnitude of displacement = 900 km. 900 km Exercise 3 Step 1: For two cars accelerating along the same road, if the situation implies the distance between them is constant (as context suggests), they must have the same acceleration. B implies that the two cars have the same acceleration ("rate"). B Exercise 4 Step 1: Object accelerates at a = 2 m/s^2 for t₁ = 5 s (assume from rest). Velocity after 5 s: v = at₁ = 2 × 5 = 10 m/s. Step 2: "Then travels 5" is ambiguous—does not specify if acceleration continues, stops, or time/distance for next phase. Cannot determine distance without more info. D Exercise 5 Step 1: Arrow projected vertically upward at initial speed u = 100 m/s, air resistance ignored (g downward). At return to ground (same level), by symmetry of motion or conservation of energy, speed v = u. v² = u² + 2as, where s=0 net displacement vertically, so v = u = 100 m/s. 100 m/s

An object undergoes constant acceleration after starting from rest and then travels 5 m in the first second. Determine how far it will go in the next second.

Answers

Step-by-step solution

Formula for average speed $v_\text{avg} = \dfrac{\text{total distance}}{\text{total time}}$ .

The aircraft flies in a straight line due north, so magnitude of displacement equals distance traveled.

For two cars accelerating along the same road, if the situation implies the distance between them is constant (as context suggests), they must have the same acceleration.

Object accelerates at $a = 2$ m/s $^2$ for $t_1 = 5$ s (assume from rest).

"Then travels 5" is ambiguous—does not specify if acceleration continues, stops, or time/distance for next phase.

Arrow projected vertically upward at initial speed $u = 100$ m/s, air resistance ignored ( $g$ downward).

Try one yourself

Still stuck on this one?

More Physics Questions

An object undergoes constant acceleration after starting from rest and then travels 5 m in the first second. Determine how far it will go in the next second.

Answers

Step-by-step solution

Formula for average speed vavg=total distancetotal timev_\text{avg} = \dfrac{\text{total distance}}{\text{total time}}vavg​=total timetotal distance​.

The aircraft flies in a straight line due north, so magnitude of displacement equals distance traveled.

For two cars accelerating along the same road, if the situation implies the distance between them is constant (as context suggests), they must have the same acceleration.

Object accelerates at a=2a = 2a=2 m/s2^22 for t1=5t_1 = 5t1​=5 s (assume from rest).

"Then travels 5" is ambiguous—does not specify if acceleration continues, stops, or time/distance for next phase.

Arrow projected vertically upward at initial speed u=100u = 100u=100 m/s, air resistance ignored (ggg downward).

Try one yourself

Still stuck on this one?

More Physics Questions

Formula for average speed $v_\text{avg} = \dfrac{\text{total distance}}{\text{total time}}$ .

Object accelerates at $a = 2$ m/s $^2$ for $t_1 = 5$ s (assume from rest).

Arrow projected vertically upward at initial speed $u = 100$ m/s, air resistance ignored ( $g$ downward).