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Baseline (Aiming for 4): List the factors

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Presentation on theme: "Baseline (Aiming for 4): List the factors"— Presentation transcript:

1 Baseline (Aiming for 4): List the factors
that affect the stopping distance of a car including braking and thinking distances. Further (Aiming for 6): Describe the relationship between speed and both thinking and braking distance. Challenge (Aiming for 8): Explain the relative effects of changes of speed on thinking and stopping distances.

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4 Example question Equations you will need: F=ma & 2as=V2-U2
1) A 800Kg ford focus can apply a force of 5000N of braking force. a) At what rate can the car negatively accelerate (decelerate) to 0m/s? b) If the car is travelling at 30m/s, what distance will it travel whilst coming to a stop? a= F/m a= 5000N/800Kg a= 6.25m/s2 2as=V2-U2 s= (V2-U2)/2a s= (02-302)/2 x 6.25 s= 72m

5 Now try these Equations you will need: F=ma & 2as=V2-U2
1) A 500Kg ford fiesta can apply a force of 4000N of braking force. a) At what rate can the car negatively accelerate (decelerate) to 0m/s? b) b) If the car is travelling at 20m/s, what distance will it travel whilst coming to a stop? 2) A 2500Kg lorry can apply a force of 6000N of braking force. a) At what rate can the lorry negatively accelerate (decelerate) to 0m/s? b) b) If the lorry is travelling at 20m/s, what distance will it travel whilst coming to a stop? 3) What effect does the increase in mass have on the braking distance?

6 Now try these Equations you will need: F=ma & 2as=V2-U2
1) A 500Kg ford fiesta can apply a force of 4000N of braking force. a) At what rate can the car negatively accelerate (decelerate) to 0m/s? b) b) If the car is travelling at 20m/s, what distance will it travel whilst coming to a stop? a= F/m a= 4000N/500Kg a= 8m/s2 2as=V2-U2 s= (V2-U2)/2a s= (02-202)/2 x 8 s= 25m

7 Now try these Equations you will need: F=ma & 2as=V2-U2
2) A 2500Kg lorry can apply a force of 6000N of braking force. a) At what rate can the lorry negatively accelerate (decelerate) to 0m/s? b) If the lorry is travelling at 20m/s, what distance will it travel whilst coming to a stop? a= F/m a= 6000N/2400Kg a= 2.5m/s2 2as=V2-U2 s= (V2-U2)/2a s= (02-202)/2 x 2.5 s= 80m

8 Fall distance / cm Reaction time / ms 1 45 26 230 51 323 76 394 2 64 27 235 52 326 77 396 3 78 28 239 53 329 399 4 90 29 243 54 332 79 402 5 101 30 247 55 335 80 404 6 111 31 252 56 338 81 407 7 119 32 256 57 341 82 409 8 128 33 260 58 344 83 412 9 135 34 263 59 347 84 414 10 143 35 267 60 350 85 416 11 150 36 271 61 353 86 419 12 156 37 275 62 356 87 421 13 163 38 278 63 359 88 424 14 169 39 282 361 89 426 15 175 40 286 65 364 429 16 181 41 289 66 367 91 431 17 186 42 293 67 370 92 433 18 192 43 296 68 373 93 436 19 197 44 300 69 375 94 438 20 202 303 70 378 95 440 21 207 46 306 71 381 96 443 22 212 47 310 72 383 97 445 23 217 48 313 73 386 98 447 24 221 49 316 74 389 99 449 25 226 50 319 75 391 100 452


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