SDT Practice SET – 3
Speed, Distance & Time: Level 3
1.
A motorist travels \(x\) percent of the total distance at 40 mph and the rest at 60 mph. If the average speed is 48 mph, what is \(x\)?
Solution (B):
Let total distance = 100.
\(x\) miles at 40 mph, \( (100-x) \) miles at 60 mph.
Total Time = \( \frac{x}{40} + \frac{100-x}{60} \).
Avg Speed = \( \frac{100}{\text{Time}} = 48 \).
\( \frac{100}{x/40 + (100-x)/60} = 48 \).
Solve for \(x\): \(x=60\).
2.
John gives Steve a 30-minute head start. John runs at 8 mph and Steve runs at 6 mph. How long will John run before he overtakes Steve?
Solution (D):
Head start dist = \( 0.5 \text{ hr} \times 6 \text{ mph} = 3 \) miles.
Relative speed = \( 8 – 6 = 2 \) mph.
Time to catch = \( 3 / 2 = 1.5 \) hours.
3.
A triathlete runs 5 miles at 10 mph and then swims 2 miles at 4 mph. What is her average speed for the entire trip (miles per hour)?
Solution (A):
Time 1 = \( 5/10 = 0.5 \) hr.
Time 2 = \( 2/4 = 0.5 \) hr.
Total Distance = 7 miles. Total Time = 1 hr.
Avg Speed = 7 mph.
4.
Two cars are 240 miles apart. Car A travels at 50 mph and Car B at 30 mph towards each other. How far from Car A’s starting point will they meet?
Solution (C):
Relative speed = \( 50 + 30 = 80 \) mph.
Time to meet = \( 240 / 80 = 3 \) hours.
Distance from A = \( 50 \times 3 = 150 \) miles.
5.
Bob bikes at \(x\) mph. He gets a flat tire exactly halfway to school. He walks the rest of the way at \(y\) mph. If the total distance is \(D\), what is his total time?
Solution (E):
First half distance = \( D/2 \). Time 1 = \( (D/2)/x = D/2x \).
Second half distance = \( D/2 \). Time 2 = \( (D/2)/y = D/2y \).
Total Time = \( \frac{D}{2x} + \frac{D}{2y} = \frac{Dy + Dx}{2xy} = \frac{D(x+y)}{2xy} \).
6.
A driver travels for 2 hours at 60 mph, then stops for 1 hour, then travels for 1 hour at 40 mph. What is the average speed for the entire period (including the stop)?
Solution (B):
Dist 1 = 120. Dist 2 = 0. Dist 3 = 40. Total Dist = 160.
Time 1 = 2. Time 2 = 1. Time 3 = 1. Total Time = 4.
Avg Speed = \( 160 / 4 = 40 \) mph.
7.
If you travel a distance at speed \(v\) and return at speed \(2v\), your average speed is what multiple of \(v\)?
Solution (D):
Avg Speed = \( \frac{2(v)(2v)}{v+2v} = \frac{4v^2}{3v} = \frac{4}{3}v = 1.33v \).
8.
At 9:00 AM, Car A leaves City X for City Y at 40 mph. At 10:00 AM, Car B leaves City X for City Y at 60 mph. At what time does Car B catch Car A?
Solution (C):
Car A head start = 1 hour \(\times\) 40 mph = 40 miles.
Relative gain = 20 mph.
Time to catch = \( 40/20 = 2 \) hours after B leaves.
10:00 AM + 2 hours = 12:00 PM.
9.
A racer runs the first lap at 40 mph. How fast must he run the second lap (same distance) to average 80 mph for both laps?
Solution (A):
Formula: \( \frac{2xy}{x+y} = 80 \). Let \(x=40\).
\( \frac{80y}{40+y} = 80 \).
Divide by 80: \( \frac{y}{40+y} = 1 \implies y = 40 + y \implies 0 = 40 \). Impossible.
Conceptually: To double the average speed over equal distances, the second leg must be done in 0 time (infinite speed).
10.
Two trains start from the same station at the same time. Train A travels North at 30 mph, Train B travels West at 40 mph. How far apart are they after 2 hours?
Solution (B):
They form a right-angled triangle.
Dist A = \(30 \times 2 = 60\).
Dist B = \(40 \times 2 = 80\).
Hypotenuse = \( \sqrt{60^2 + 80^2} = \sqrt{3600 + 6400} = \sqrt{10000} = 100 \).
Score: 0 / 0