A car traveling 90km/hr is 100 m behind a truck traveling 50km/hr. How long will it take the car to reach the truck?

2 answers:

5 0

Since both these cars are traveling as an assumed constant rate, you can imagine them as not moving at all to make the problem easier conceptually. The distance between the cars will also remain constant. First, convert the units km/hr into m/s: 90/3.6= 25 m/s; 50/3.6=13.89 m/s. The difference between these two speeds is 11.11 m/s. Now use the equation: velocity= distance/time. 11.11= 100/time; solve for time and you should get 9.0009 seconds

AlexFokin [52]4 years ago

4 0

The faster car behind is catching up/closing the gap/gaining on
the slow truck in front at the rate of (90 - 50) = 40 km/hr.

At that rate, it takes (100 m) / (40,000 m/hr) = 1/400 of an hour
to reach the truck.

(1/400 hour) x (3,600 seconds/hour) = 3600/400 = <em>9 seconds</em>, exactly

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Sonja [21]

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Salmon often jump waterfalls to reach their breeding grounds. Starting downstream, 3.04 m away from a waterfall 0.585 m in heigh

S_A_V [24]

Answer:

V₀ = 5.47 m/s

Explanation:

The jumping motion of the Salmon can be modelled as the projectile motion. So, we use the formula for the range of projectile motion here:

R = V₀² Sin 2θ/g

where,

R = Range of Projectile = 3.04 m

θ = Launch Angle = 41.7°

V₀ = Minimum Launch Speed = ?

g = 9.81 m/s²

Therefore,

3.04 m = V₀² [Sin2(41.7°)]/(9.81 m/s²)

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3 years ago

A 26 foot ladder is lowered down a vertical wall at a rate of 3 feet per minute. The base of the ladder is sliding away from the

lakkis [162]

Answer:

(i) 7.2 feet per minute.

(ii) No, the rate would be different.

(iii) The rate would be always positive.

(iv) the resultant change would be constant.

(v) 0 feet per min

Explanation:

Let the length of ladder is l, x be the height of the top of the ladder from the ground and y be the length of the bottom of the ladder from the wall,

By making the diagram of this situation,

Applying Pythagoras theorem,

$l^2 = x^2 + y^2-----(1)$